Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 4 Expansions Including Substitution

Exercise 4(A)

Question 1. Find the square of:
(i) \( 2a + b \)
(ii) \( 3a + 7b \)
(iii) \( 3a - 4b \)
(iv) \( \frac{3a}{2b} - \frac{2b}{3a} \)
Answer:
(i) Applying the standard algebraic identity \( (x + y)^2 = x^2 + 2xy + y^2 \):
\( (2a + b)^2 = (2a)^2 + 2(2a)(b) + (b)^2 \)
\( = 4a^2 + 4ab + b^2 \)

(ii) Applying the same expansion identity for a sum:
\( (3a + 7b)^2 = (3a)^2 + 2(3a)(7b) + (7b)^2 \)
\( = 9a^2 + 42ab + 49b^2 \)

(iii) Applying the standard identity for a difference \( (x - y)^2 = x^2 - 2xy + y^2 \):
\( (3a - 4b)^2 = (3a)^2 - 2(3a)(4b) + (4b)^2 \)
\( = 9a^2 - 24ab + 16b^2 \)

(iv) Applying the difference identity to fractional terms:
\( \left(\frac{3a}{2b} - \frac{2b}{3a}\right)^2 = \left(\frac{3a}{2b}\right)^2 - 2\left(\frac{3a}{2b}\right)\left(\frac{2b}{3a}\right) + \left(\frac{2b}{3a}\right)^2 \)
\( = \frac{9a^2}{4b^2} - 2 + \frac{4b^2}{9a^2} \)
In simple words: To square these expressions, we use the standard algebraic formulas \( (a+b)^2 = a^2 + b^2 + 2ab \) and \( (a-b)^2 = a^2 + b^2 - 2ab \). Substitute the terms into the formula and simplify.

Exam Tip: Pay close attention to signs when expanding, especially with negative terms. Ensure you square both the numerical coefficients and the variables inside parentheses.

Question 2. Use identities to evaluate:
(i) \( (101)^2 \)
(ii) \( (502)^2 \)
(iii) \( (97)^2 \)
(iv) \( (998)^2 \)
Answer:
(i) Express \( 101 \) as \( 100 + 1 \). Expand using \( (x+y)^2 = x^2 + 2xy + y^2 \):
\( (101)^2 = (100 + 1)^2 \)
\( = (100)^2 + 2(100)(1) + (1)^2 \)
\( = 10000 + 200 + 1 \)
\( = 10201 \)

(ii) Express \( 502 \) as \( 500 + 2 \). Expand using \( (x+y)^2 = x^2 + 2xy + y^2 \):
\( (502)^2 = (500 + 2)^2 \)
\( = (500)^2 + 2(500)(2) + (2)^2 \)
\( = 250000 + 2000 + 4 \)
\( = 252004 \)

(iii) Express \( 97 \) as \( 100 - 3 \). Expand using \( (x-y)^2 = x^2 - 2xy + y^2 \):
\( (97)^2 = (100 - 3)^2 \)
\( = (100)^2 - 2(100)(3) + (3)^2 \)
\( = 10000 - 600 + 9 \)
\( = 9409 \)

(iv) Express \( 998 \) as \( 1000 - 2 \). Expand using \( (x-y)^2 = x^2 - 2xy + y^2 \):
\( (998)^2 = (1000 - 2)^2 \)
\( = (1000)^2 - 2(1000)(2) + (2)^2 \)
\( = 1000000 - 4000 + 4 \)
\( = 996004 \)
In simple words: Instead of multiplying these large numbers directly, split them into base numbers like 100, 500, or 1000 plus or minus a small number. Then use the standard squared identities to make the calculations easier.

Exam Tip: Never calculate the squares by direct multiplication. The question specifically asks you to "use identities," so you must show the expansion steps to receive full credit.

Question 3. Evaluate:
(i) \( \left(\frac{7}{8}x + \frac{4}{5}y\right)^2 \)
(ii) \( \left(\frac{2x}{7} - \frac{7y}{4}\right)^2 \)
Answer:
(i) Expand the term \( \left(\frac{7}{8}x + \frac{4}{5}y\right)^2 \) using the standard binomial identity \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( \left(\frac{7}{8}x + \frac{4}{5}y\right)^2 = \left(\frac{7}{8}x\right)^2 + 2\left(\frac{7}{8}x\right)\left(\frac{4}{5}y\right) + \left(\frac{4}{5}y\right)^2 \)
\( = \frac{49x^2}{64} + \frac{7xy}{5} + \frac{16y^2}{25} \)

(ii) Expand the term \( \left(\frac{2x}{7} - \frac{7y}{4}\right)^2 \) using the identity \( (a-b)^2 = a^2 - 2ab + b^2 \):
\( \left(\frac{2x}{7} - \frac{7y}{4}\right)^2 = \left(\frac{2x}{7}\right)^2 - 2\left(\frac{2x}{7}\right)\left(\frac{7y}{4}\right) + \left(\frac{7y}{4}\right)^2 \)
\( = \frac{4x^2}{49} - xy + \frac{49y^2}{16} \)
In simple words: Squaring terms with fractions works the same way. Square each numerator and denominator, and when expanding the middle term, cancel common terms to simplify the final fraction.

Exam Tip: When multiplying the fractional middle term, always look for common factors to cancel out (for example, the 2 and 8 in part i, and the 7s and 2, 4 in part ii) to avoid leaving unsimplified fractions.

Question 4. Evaluate:
(i) \( \left(\frac{a}{2b} + \frac{2b}{a}\right)^2 - \left(\frac{a}{2b} - \frac{2b}{a}\right)^2 - 4 \)
(ii) \( (4a + 3b)^2 - (4a - 3b)^2 + 48ab \)
Answer:
(i) First, let's expand the positive term \( \left(\frac{a}{2b} + \frac{2b}{a}\right)^2 \) using the sum identity:
\( \left(\frac{a}{2b} + \frac{2b}{a}\right)^2 = \left(\frac{a}{2b}\right)^2 + \left(\frac{2b}{a}\right)^2 + 2\left(\frac{a}{2b}\right)\left(\frac{2b}{a}\right) \)
\( = \frac{a^2}{4b^2} + \frac{4b^2}{a^2} + 2 \quad \text{--- (1)} \)

Now, let's expand the negative term \( \left(\frac{a}{2b} - \frac{2b}{a}\right)^2 \) using the difference identity:
\( \left(\frac{a}{2b} - \frac{2b}{a}\right)^2 = \left(\frac{a}{2b}\right)^2 + \left(\frac{2b}{a}\right)^2 - 2\left(\frac{a}{2b}\right)\left(\frac{2b}{a}\right) \)
\( = \frac{a^2}{4b^2} + \frac{4b^2}{a^2} - 2 \quad \text{--- (2)} \)

Substitute equations (1) and (2) back into the main expression:
\( \left(\frac{a}{2b} + \frac{2b}{a}\right)^2 - \left(\frac{a}{2b} - \frac{2b}{a}\right)^2 - 4 = \left(\frac{a^2}{4b^2} + \frac{4b^2}{a^2} + 2\right) - \left(\frac{a^2}{4b^2} + \frac{4b^2}{a^2} - 2\right) - 4 \)
\( = \frac{a^2}{4b^2} + \frac{4b^2}{a^2} + 2 - \frac{a^2}{4b^2} - \frac{4b^2}{a^2} + 2 - 4 \)
\( = 4 - 4 = 0 \)

(ii) Expand the first term \( (4a + 3b)^2 \):
\( (4a + 3b)^2 = (4a)^2 + (3b)^2 + 2(4a)(3b) \)
\( = 16a^2 + 9b^2 + 24ab \quad \text{--- (1)} \)

Next, expand the second term \( (4a - 3b)^2 \):
\( (4a - 3b)^2 = (4a)^2 + (3b)^2 - 2(4a)(3b) \)
\( = 16a^2 + 9b^2 - 24ab \quad \text{--- (2)} \)

Combine these expansions as indicated in the given expression:
\( (4a + 3b)^2 - (4a - 3b)^2 + 48ab = (16a^2 + 9b^2 + 24ab) - (16a^2 + 9b^2 - 24ab) + 48ab \)
\( = 16a^2 + 9b^2 + 24ab - 16a^2 - 9b^2 + 24ab + 48ab \)
\( = 24ab + 24ab + 48ab \)
\( = 96ab \)
In simple words: When subtracting one squared binomial from another, most of the squared terms cancel out, leaving behind only the middle terms. Group the like terms together to simplify the final answer.

Exam Tip: Be very careful with the negative sign outside the second parentheses. When you remove the brackets, it changes the signs of all terms inside (making the positive terms negative and the negative term positive).

Question 5. If \( a + b = 7 \) and \( ab = 10 \); find \( a - b \).
Answer:
We can relate the sum, difference, and product of two variables using their squared identities:
\( (a - b)^2 = (a + b)^2 - 4ab \quad \text{--- (1)} \)

Substitute the given values \( a + b = 7 \) and \( ab = 10 \) into equation (1):
\( (a - b)^2 = (7)^2 - 4(10) \)
\( = 49 - 40 \)
\( = 9 \)

Taking the square root on both sides:
\( a - b = \pm \sqrt{9} \)
\( = \pm 3 \)
In simple words: We can find the difference between two numbers directly using the identity \( (a-b)^2 = (a+b)^2 - 4ab \) without needing to find the individual values of \( a \) and \( b \) first.

Exam Tip: Always remember to write \( \pm \) when taking the square root of a number, as both positive and negative values squared yield the same positive result.

Question 6. If \( a - b = 7 \) and \( ab = 18 \); find \( a + b \).
Answer:
The relation between the sum, difference, and product of two terms is given by the identity:
\( (a + b)^2 = (a - b)^2 + 4ab \quad \text{--- (1)} \)

Substitute the given values \( a - b = 7 \) and \( ab = 18 \) into equation (1):
\( (a + b)^2 = (7)^2 + 4(18) \)
\( = 49 + 72 \)
\( = 121 \)

Taking the square root on both sides:
\( a + b = \pm \sqrt{121} \)
\( = \pm 11 \)
In simple words: This uses the formula \( (a+b)^2 = (a-b)^2 + 4ab \). Plugging in the difference and product helps us calculate the sum directly.

Exam Tip: Show the formula clearly before substituting the values to ensure step-marks are awarded even if a minor arithmetic error occurs later.

Question 7. If \( x + y = \frac{7}{2} \) and \( xy = \frac{5}{2} \); find:
(i) \( x - y \)
(ii) \( x^2 - y^2 \)
Answer:
(i) We use the algebraic relation between sum and difference:
\( (x - y)^2 = (x + y)^2 - 4xy \quad \text{--- (1)} \)

Substitute \( x + y = \frac{7}{2} \) and \( xy = \frac{5}{2} \) into equation (1):
\( (x - y)^2 = \left(\frac{7}{2}\right)^2 - 4\left(\frac{5}{2}\right) \)
\( = \frac{49}{4} - 10 \)
\( = \frac{49 - 40}{4} \)
\( = \frac{9}{4} \)

Taking the square root of both sides gives:
\( x - y = \pm \sqrt{\frac{9}{4}} \)
\( = \pm \frac{3}{2} \)

(ii) We apply the difference of squares identity:
\( x^2 - y^2 = (x + y)(x - y) \quad \text{--- (2)} \)

Using the values \( x + y = \frac{7}{2} \) and \( x - y = \pm \frac{3}{2} \) in equation (2):
\( x^2 - y^2 = \left(\frac{7}{2}\right) \left(\pm \frac{3}{2}\right) \)
\( = \pm \frac{21}{4} \)
In simple words: First, find the difference \( x - y \) using the sum and product formula. Once you have both the sum and the difference, multiply them together to get \( x^2 - y^2 \).

Exam Tip: Since \( x - y \) has two possible values (\( \pm \frac{3}{2} \)), your final answer for \( x^2 - y^2 \) must also reflect both positive and negative results (\( \pm \frac{21}{4} \)).

Question 8. If \( a - b = 0.9 \) and \( ab = 0.36 \); find:
(i) \( a + b \)
(ii) \( a^2 - b^2 \)
Answer:
(i) Apply the identity relating sum, difference, and product:
\( (a + b)^2 = (a - b)^2 + 4ab \quad \text{--- (1)} \)

Substitute \( a - b = 0.9 \) and \( ab = 0.36 \) into equation (1):
\( (a + b)^2 = (0.9)^2 + 4(0.36) \)
\( = 0.81 + 1.44 \)
\( = 2.25 \)

Taking the square root on both sides:
\( a + b = \pm \sqrt{2.25} \)
\( = \pm 1.5 \)

(ii) The difference of squares is defined as:
\( a^2 - b^2 = (a + b)(a - b) \quad \text{--- (2)} \)

Substitute \( a + b = \pm 1.5 \) and \( a - b = 0.9 \) into equation (2):
\( a^2 - b^2 = (\pm 1.5)(0.9) \)
\( = \pm 1.35 \)
In simple words: This problem follows the same steps but with decimals. Calculate the sum first using the relation \( (a+b)^2 = (a-b)^2 + 4ab \), then multiply the sum and difference to find the final value.

Exam Tip: Work carefully with decimal multiplication and addition to avoid simple calculation errors, such as misplacing the decimal point in \( 0.9^2 = 0.81 \) or \( 4 \times 0.36 = 1.44 \).

Question 9. If \( a + b = 6 \) and \( a - b = 4 \); find:
(i) \( a^2 + b^2 \)
(ii) \( ab \)
Answer:
(i) Consider the two expansion formulas:
\( (a + b)^2 = a^2 + b^2 + 2ab \quad \text{--- (1)} \)
\( (a - b)^2 = a^2 + b^2 - 2ab \quad \text{--- (2)} \)

Adding equations (1) and (2) gives:
\( (a + b)^2 + (a - b)^2 = 2(a^2 + b^2) \)
\( \implies 2(a^2 + b^2) = (a + b)^2 + (a - b)^2 \)
\( \implies a^2 + b^2 = \frac{1}{2}\left[(a + b)^2 + (a - b)^2\right] \quad \text{--- (3)} \)

Substitute \( a + b = 6 \) and \( a - b = 4 \) into equation (3):
\( a^2 + b^2 = \frac{1}{2}\left[(6)^2 + (4)^2\right] \)
\( = \frac{1}{2}[36 + 16] \)
\( = \frac{52}{2} \)
\( = 26 \)

(ii) To find \( ab \), substitute \( a^2 + b^2 = 26 \) and \( a - b = 4 \) into equation (2):
\( (a - b)^2 = a^2 + b^2 - 2ab \)
\( \implies (4)^2 = 26 - 2ab \)
\( \implies 16 = 26 - 2ab \)
\( \implies 2ab = 26 - 16 \)
\( \implies 2ab = 10 \)
\( \implies ab = 5 \)
In simple words: By adding the squared sum and squared difference, we can find \( a^2+b^2 \). After finding \( a^2+b^2 \), we can substitute it back into one of the squared formulas to solve for \( ab \).

Exam Tip: A useful shortcut for part (ii) is using the identity \( (a+b)^2 - (a-b)^2 = 4ab \). This avoids using the intermediate \( a^2+b^2 \) value entirely and can be a great way to double-check your answer.

Question 10. If \( a + \frac{1}{a} = 6 \); find:
(i) \( a - \frac{1}{a} \)
(ii) \( a^2 - \frac{1}{a^2} \)
Answer:
(i) We know that:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \quad \text{--- (1)} \)

Substitute \( a + \frac{1}{a} = 6 \) into equation (1):
\( (6)^2 = a^2 + \frac{1}{a^2} + 2 \quad \)
\( \implies 36 = a^2 + \frac{1}{a^2} + 2 \)
\( \implies a^2 + \frac{1}{a^2} = 34 \quad \text{--- (2)} \)

Now, let's use the identity for subtraction:
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \)

Using the value from equation (2):
\( \left(a - \frac{1}{a}\right)^2 = 34 - 2 \)
\( = 32 \)

Taking the square root on both sides:
\( a - \frac{1}{a} = \pm \sqrt{32} \)
\( \implies a - \frac{1}{a} = \pm 4\sqrt{2} \quad \text{--- (3)} \)

(ii) To find the value of \( a^2 - \frac{1}{a^2} \), we apply the difference of squares:
\( a^2 - \frac{1}{a^2} = \left(a + \frac{1}{a}\right)\left(a - \frac{1}{a}\right) \)

Substitute the values \( a + \frac{1}{a} = 6 \) and \( a - \frac{1}{a} = \pm 4\sqrt{2} \):
\( a^2 - \frac{1}{a^2} = (6)\left(\pm 4\sqrt{2}\right) \)
\( \implies a^2 - \frac{1}{a^2} = \pm 24\sqrt{2} \)
In simple words: First, find the value of \( a^2 + \frac{1}{a^2} \). Use that intermediate value to calculate \( a - \frac{1}{a} \). Finally, multiply the sum and the difference terms together to find \( a^2 - \frac{1}{a^2} \).

Exam Tip: When simplifying square roots like \( \sqrt{32} \), always factor out the largest perfect square factor (which is 16 here, so \( \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \)) for a fully simplified final answer.

Question 11. If \( a - \frac{1}{a} = 8 \); find:
(i) \( a + \frac{1}{a} \)
(ii) \( a^2 - \frac{1}{a^2} \)
Answer:
(i) We know that:
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \quad \text{--- (1)} \)

Substitute \( a - \frac{1}{a} = 8 \) into equation (1):
\( (8)^2 = a^2 + \frac{1}{a^2} - 2 \)
\( \implies 64 = a^2 + \frac{1}{a^2} - 2 \)
\( \implies a^2 + \frac{1}{a^2} = 66 \quad \text{--- (2)} \)

Now, let's use the positive squared identity:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)

Using the value from equation (2):
\( \left(a + \frac{1}{a}\right)^2 = 66 + 2 \)
\( = 68 \)

Taking the square root on both sides:
\( a + \frac{1}{a} = \pm \sqrt{68} \)
\( \implies a + \frac{1}{a} = \pm 2\sqrt{17} \quad \text{--- (3)} \)

(ii) Applying the difference of squares identity:
\( a^2 - \frac{1}{a^2} = \left(a - \frac{1}{a}\right)\left(a + \frac{1}{a}\right) \)

Substitute the values \( a - \frac{1}{a} = 8 \) and \( a + \frac{1}{a} = \pm 2\sqrt{17} \):
\( a^2 - \frac{1}{a^2} = (8)\left(\pm 2\sqrt{17}\right) \)
\( \implies a^2 - \frac{1}{a^2} = \pm 16\sqrt{17} \)
In simple words: This problem is similar to the previous one but in reverse. Find the sum \( a^2 + \frac{1}{a^2} \) first, then find \( a + \frac{1}{a} \), and finally multiply the terms to get \( a^2 - \frac{1}{a^2} \).

Exam Tip: Be careful when simplifying radical terms like \( \sqrt{68} \). Since \( 68 = 4 \times 17 \), the square root simplifies to \( 2\sqrt{17} \).

Question 12. If \( a^2 - 3a + 1 = 0 \); find:
(i) \( a + \frac{1}{a} \)
(ii) \( a^2 + \frac{1}{a^2} \)
Answer:
(i) Let's start with the given equation:
\( a^2 - 3a + 1 = 0 \)

Rearranging the terms:
\( a^2 + 1 = 3a \)

Dividing both sides of the equation by \( a \):
\( \frac{a^2 + 1}{a} = 3 \)
\( \implies \frac{a^2}{a} + \frac{1}{a} = 3 \)
\( \implies a + \frac{1}{a} = 3 \quad \text{--- (1)} \)

(ii) To find the value of \( a^2 + \frac{1}{a^2} \), we use the identity:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \quad \text{--- (2)} \)

Using the value \( a + \frac{1}{a} = 3 \) from equation (1) in equation (2):
\( (3)^2 = a^2 + \frac{1}{a^2} + 2 \)
\( \implies 9 = a^2 + \frac{1}{a^2} + 2 \)
\( \implies a^2 + \frac{1}{a^2} = 9 - 2 \)
\( \implies a^2 + \frac{1}{a^2} = 7 \)
In simple words: First, divide the whole quadratic equation by \( a \) to convert it into a simple fractional expression \( a + \frac{1}{a} \). Once you have that, square it to get \( a^2 + \frac{1}{a^2} \).

Exam Tip: Dividing a quadratic equation of the form \( x^2 - kx + 1 = 0 \) by \( x \) is a very common algebraic trick that instantly gives \( x + \frac{1}{x} = k \). Memorize this technique!

Question 13. If \( a^2 - 5a - 1 = 0 \); find:
(i) \( a - \frac{1}{a} \)
(ii) \( a + \frac{1}{a} \)
Answer:
(i) Consider the given quadratic equation:
\( a^2 - 5a - 1 = 0 \)

Rearrange the terms:
\( a^2 - 1 = 5a \)

Divide the entire equation by \( a \):
\( \frac{a^2 - 1}{a} = 5 \)
\( \implies \frac{a^2}{a} - \frac{1}{a} = 5 \)
\( \implies a - \frac{1}{a} = 5 \quad \text{--- (1)} \)

(ii) We know the identity relating \( a + \frac{1}{a} \) and \( a - \frac{1}{a} \):
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \)

Substitute \( a - \frac{1}{a} = 5 \) from equation (1):
\( (5)^2 = a^2 + \frac{1}{a^2} - 2 \)
\( \implies 25 = a^2 + \frac{1}{a^2} - 2 \)
\( \implies a^2 + \frac{1}{a^2} = 27 \quad \text{--- (2)} \)

Now, to find the value of \( a + \frac{1}{a} \), consider the positive squared identity:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)

Using the value from equation (2):
\( \left(a + \frac{1}{a}\right)^2 = 27 + 2 \)
\( \implies \left(a + \frac{1}{a}\right)^2 = 29 \)

Taking the square root on both sides:
\( a + \frac{1}{a} = \pm \sqrt{29} \)
In simple words: Divide the quadratic equation by \( a \) to get the subtraction value \( a - \frac{1}{a} \). Use this to calculate the squared term \( a^2 + \frac{1}{a^2} \), and finally use that to find \( a + \frac{1}{a} \).

Exam Tip: Since \( 29 \) is a prime number, leave \( \pm\sqrt{29} \) in its radical form rather than calculating its decimal value.

Question 14. If \( 3x + 4y = 16 \) and \( xy = 4 \); find the value of \( 9x^2 + 16y^2 \).
Answer:
We are given:
\( 3x + 4y = 16 \) and \( xy = 4 \)

Let's square the first expression:
\( (3x + 4y)^2 = (3x)^2 + (4y)^2 + 2(3x)(4y) \)
\( \implies (3x + 4y)^2 = 9x^2 + 16y^2 + 24xy \quad \text{--- (1)} \)

Now, substitute the values \( 3x + 4y = 16 \) and \( xy = 4 \) into equation (1):
\( (16)^2 = 9x^2 + 16y^2 + 24(4) \)
\( \implies 256 = 9x^2 + 16y^2 + 96 \)
\( \implies 9x^2 + 16y^2 = 256 - 96 \)
\( \implies 9x^2 + 16y^2 = 160 \)
In simple words: Square the given equation \( 3x + 4y = 16 \). Then replace \( xy \) with 4 and simplify the arithmetic to find the value of \( 9x^2 + 16y^2 \).

Exam Tip: When squaring the term \( (3x + 4y) \), remember that the middle term is \( 2 \times 3x \times 4y = 24xy \). A very common error is writing \( 12xy \) instead of \( 24xy \).

Question 15. The number \( x \) is \( 2 \) more than the number \( y \). If the sum of the squares of \( x \) and \( y \) is \( 34 \), find the product of \( x \) and \( y \).
Answer:
According to the first condition, \( x \) is 2 greater than \( y \):
\( x = y + 2 \quad \text{--- (1)} \)

According to the second condition, the sum of their squares is 34:
\( x^2 + y^2 = 34 \quad \text{--- (2)} \)

Substitute equation (1) into equation (2):
\( (y + 2)^2 + y^2 = 34 \)
\( \implies (y^2 + 4y + 4) + y^2 = 34 \)
\( \implies 2y^2 + 4y + 4 = 34 \)
\( \implies 2y^2 + 4y + 4 - 34 = 0 \)
\( \implies 2y^2 + 4y - 30 = 0 \)

Dividing the entire quadratic equation by 2:
\( y^2 + 2y - 15 = 0 \)

Factorise this quadratic equation by splitting the middle term:
\( y^2 + 5y - 3y - 15 = 0 \)
\( \implies y(y + 5) - 3(y + 5) = 0 \)
\( \implies (y + 5)(y - 3) = 0 \)

So, we have:
\( y = -5 \quad \text{or} \quad y = 3 \)

Let's find the corresponding values of \( x \) for both cases:

Case 1: If \( y = -5 \), then:
\( x = -5 + 2 = -3 \)
The product of the numbers is:
\( xy = (-3)(-5) = 15 \)

Case 2: If \( y = 3 \), then:
\( x = 3 + 2 = 5 \)
The product of the numbers is:
\( xy = (5)(3) = 15 \)

Thus, in both scenarios, the product of \( x \) and \( y \) is 15.
In simple words: Create equations from the word problem, solve the resulting quadratic equation to find the two possible sets of values for \( x \) and \( y \), and multiply them to get the same product of 15.

Exam Tip: Be sure to write down and evaluate both possible algebraic cases (Case 1 where numbers are negative, and Case 2 where they are positive) to ensure you do not lose marks for an incomplete solution.

Question 16. The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.
Answer:
Assume the two positive values are \( a \) and \( b \).
As per the given details:
\( a - b = 5 \) and \( a^2 + b^2 = 73 \)
We square both sides of the subtraction equation:
\( (a - b)^2 = 5^2 \)
\( \implies a^2 + b^2 - 2ab = 25 \)
Since \( a^2 + b^2 = 73 \), we get:
\( 73 - 2ab = 25 \)
\( \implies 2ab = 73 - 25 \)
\( \implies 2ab = 48 \)
\( \implies ab = 24 \)
Thus, the product of these numbers equals 24.
In simple words: If you know the difference between two numbers and the sum of their squares, you can use a basic algebraic identity to find their product.

Exam Tip: Be careful with the signs when moving terms from one side of the equation to the other.

Exercise 4(B)

Question 1. Expand the following expressions:
(i) \( (3a - 2b)^3 \)
(ii) \( (5a + 3b)^3 \)
(iii) \( \left(2a + \frac{1}{2a}\right)^3 \)
(iv) \( \left(3a - \frac{1}{a}\right)^3 \)
Answer:
(i) Using the identity \( (x - y)^3 = x^3 - 3xy(x - y) - y^3 \), we expand \( (3a - 2b)^3 \):
\( (3a - 2b)^3 = (3a)^3 - 3(3a)(2b)(3a - 2b) - (2b)^3 \)
\( \implies = 27a^3 - 18ab(3a - 2b) - 8b^3 \)
\( \implies = 27a^3 - 54a^2b + 36ab^2 - 8b^3 \)

(ii) Using the identity \( (x + y)^3 = x^3 + 3xy(x + y) + y^3 \), we expand \( (5a + 3b)^3 \):
\( (5a + 3b)^3 = (5a)^3 + 3(5a)(3b)(5a + 3b) + (3b)^3 \)
\( \implies = 125a^3 + 45ab(5a + 3b) + 27b^3 \)
\( \implies = 125a^3 + 225a^2b + 135ab^2 + 27b^3 \)

(iii) Expanding \( \left(2a + \frac{1}{2a}\right)^3 \) using the sum of cubes identity:
\( \left(2a + \frac{1}{2a}\right)^3 = (2a)^3 + 3(2a)\left(\frac{1}{2a}\right)\left(2a + \frac{1}{2a}\right) + \left(\frac{1}{2a}\right)^3 \)
\( \implies = 8a^3 + 3\left(2a + \frac{1}{2a}\right) + \frac{1}{8a^3} \)
\( \implies = 8a^3 + 6a + \frac{3}{2a} + \frac{1}{8a^3} \)

(iv) Expanding \( \left(3a - \frac{1}{a}\right)^3 \) using the difference of cubes identity:
\( \left(3a - \frac{1}{a}\right)^3 = (3a)^3 - 3(3a)\left(\frac{1}{a}\right)\left(3a - \frac{1}{a}\right) - \left(\frac{1}{a}\right)^3 \)
\( \implies = 27a^3 - 9\left(3a - \frac{1}{a}\right) - \frac{1}{a^3} \)
\( \implies = 27a^3 - 27a + \frac{9}{a} - \frac{1}{a^3} \)
In simple words: To expand cubic brackets, use the standard formulas for \( (a+b)^3 \) or \( (a-b)^3 \) and carefully simplify each term step-by-step.

Exam Tip: Always keep track of signs when expanding \( (a-b)^3 \), particularly when multiplying a negative term across the terms inside.

Question 2. If \( a^2 + \frac{1}{a^2} = 47 \), find the value of:
(i) \( a + \frac{1}{a} \)
(ii) \( a^3 + \frac{1}{a^3} \)
Answer:
(i) We know that:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)
Substitute the given value:
\( \left(a + \frac{1}{a}\right)^2 = 47 + 2 \)
\( \implies \left(a + \frac{1}{a}\right)^2 = 49 \)
Taking square roots on both sides:
\( \implies a + \frac{1}{a} = \pm\sqrt{49} \)
\( \implies a + \frac{1}{a} = \pm 7 \) ....(1)

(ii) We know that:
\( \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
Rearranging the terms:
\( \implies a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right) \)
Using equation (1), substitute the values of \( a + \frac{1}{a} \):
\( \implies a^3 + \frac{1}{a^3} = (\pm 7)^3 - 3(\pm 7) \)
This gives us two cases:
Case 1: When \( a + \frac{1}{a} = 7 \)
\( a^3 + \frac{1}{a^3} = (7)^3 - 3(7) = 343 - 21 = 322 \)
Case 2: When \( a + \frac{1}{a} = -7 \)
\( a^3 + \frac{1}{a^3} = (-7)^3 - 3(-7) = -343 + 21 = -322 \)
Thus, combining both cases:
\( \implies a^3 + \frac{1}{a^3} = \pm 322 \)
In simple words: We can find the value of a sum of cubes or a sum of squares by building them out of the basic sum term \( a + \frac{1}{a} \).

Exam Tip: When taking a square root, remember to include both the positive and negative signs (\( \pm \)), as they both yield valid algebraic solutions.

Question 3. If \( a^2 + \frac{1}{a^2} = 18 \), find the value of:
(i) \( a - \frac{1}{a} \)
(ii) \( a^3 - \frac{1}{a^3} \)
Answer:
(i) We have the identity:
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \)
Substituting the given value:
\( \left(a - \frac{1}{a}\right)^2 = 18 - 2 \)
\( \implies \left(a - \frac{1}{a}\right)^2 = 16 \)
\( \implies a - \frac{1}{a} = \pm\sqrt{16} \)
\( \implies a - \frac{1}{a} = \pm 4 \) ....(1)

(ii) Using the identity for the cube of a difference:
\( \left(a - \frac{1}{a}\right)^3 = a^3 - \frac{1}{a^3} - 3\left(a - \frac{1}{a}\right) \)
Rearranging the terms:
\( \implies a^3 - \frac{1}{a^3} = \left(a - \frac{1}{a}\right)^3 + 3\left(a - \frac{1}{a}\right) \)
Substitute the value from equation (1):
\( \implies a^3 - \frac{1}{a^3} = (\pm 4)^3 + 3(\pm 4) \)
This gives us two cases:
Case 1: When \( a - \frac{1}{a} = 4 \)
\( a^3 - \frac{1}{a^3} = 4^3 + 3(4) = 64 + 12 = 76 \)
Case 2: When \( a - \frac{1}{a} = -4 \)
\( a^3 - \frac{1}{a^3} = (-4)^3 + 3(-4) = -64 - 12 = -76 \)
Thus, combining both cases:
\( \implies a^3 - \frac{1}{a^3} = \pm 76 \)
In simple words: First find the value of the difference \( a - \frac{1}{a} \) by using the squared identity, and then use that result in the cubic formula.

Exam Tip: Be extra careful with the sign in front of the 3 in the cube formulas: for \( a^3 - \frac{1}{a^3} \), the term added is \( +3\left(a - \frac{1}{a}\right) \).

Question 4. If \( a + \frac{1}{a} = p \), show that \( a^3 + \frac{1}{a^3} = p(p^2 - 3) \).
Answer:
We are given:
\( a + \frac{1}{a} = p \) ....(1)
Consider the identity:
\( \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
By rearranging this equation, we get:
\( \implies a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right) \)
Substitute the given value from equation (1):
\( \implies a^3 + \frac{1}{a^3} = (p)^3 - 3(p) \)
Factoring out \( p \) from both terms:
\( \implies a^3 + \frac{1}{a^3} = p(p^2 - 3) \)
Hence proved.
In simple words: We can write the sum of cubes in terms of the variable \( p \) by substituting \( p \) into the expanded cube identity and factoring.

Exam Tip: Factoring out the common variable in the final step is essential to get the exact algebraic form asked in the proof.

Question 5. If \( a + 2b = 5 \), find the value of \( a^3 + 8b^3 + 30ab \).
Answer:
Given:
\( a + 2b = 5 \)
We need to find the value of the expression \( a^3 + 8b^3 + 30ab \).
Let us take the cube of the given relation \( a + 2b = 5 \):
\( (a + 2b)^3 = a^3 + (2b)^3 + 3 \times a \times 2b \times (a + 2b) \)
\( \implies (a + 2b)^3 = a^3 + 8b^3 + 6ab(a + 2b) \)
Substitute the given value \( a + 2b = 5 \):
\( 5^3 = a^3 + 8b^3 + 6ab(5) \)
\( \implies 125 = a^3 + 8b^3 + 30ab \)
Thus, the value of \( a^3 + 8b^3 + 30ab \) is 125.
In simple words: Cubing the basic expression \( a + 2b \) gives us a formula that contains the target expression. We then substitute the given numerical value to find the answer.

Exam Tip: Recognize that \( 8b^3 \) is the cube of \( 2b \). This signals that cubing both sides of the initial linear equation is the correct path.

Question 6. If \( \left(a + \frac{1}{a}\right)^2 = 3 \), find the value of \( a^3 + \frac{1}{a^3} \).
Answer:
Given that:
\( \left(a + \frac{1}{a}\right)^2 = 3 \)
Taking the square root on both sides:
\( \implies a + \frac{1}{a} = \pm\sqrt{3} \) ....(1)
We need to find the value of \( a^3 + \frac{1}{a^3} \).
Consider the identity:
\( \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
Rearranging the formula:
\( \implies a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right) \)
From (1), substitute \( a + \frac{1}{a} = \pm\sqrt{3} \):
\( \implies a^3 + \frac{1}{a^3} = (\pm\sqrt{3})^3 - 3(\pm\sqrt{3}) \)
\( \implies a^3 + \frac{1}{a^3} = \pm 3\sqrt{3} - 3(\pm\sqrt{3}) \)
\( \implies a^3 + \frac{1}{a^3} = 0 \)
Therefore, the required value is 0.
In simple words: First, take the square root of the given equation. Then, use the cubed formula where the terms will cancel each other out, leaving a final answer of 0.

Exam Tip: Note that \( (\sqrt{3})^3 = 3\sqrt{3} \). This exact cancellation is a common pattern in algebraic simplifications.

Question 7. If \( a + 2b + c = 0 \), show that \( a^3 + 8b^3 + c^3 = 6abc \).
Answer:
Given that:
\( a + 2b + c = 0 \)
\( \implies a + 2b = -c \) ....(1)
Now, let us find the cube of both sides of this equation:
\( (a + 2b)^3 = (-c)^3 \)
\( \implies a^3 + (2b)^3 + 3 \times a \times 2b \times (a + 2b) = -c^3 \)
Using the relation from (1), we replace \( a + 2b \) with \( -c \):
\( \implies a^3 + 8b^3 + 6ab(-c) = -c^3 \)
\( \implies a^3 + 8b^3 - 6abc = -c^3 \)
Move the negative terms to rearrange:
\( \implies a^3 + 8b^3 + c^3 = 6abc \)
Hence proved.
In simple words: Move \( c \) to the other side to get \( a + 2b = -c \). Cube both sides and substitute \( -c \) back in to get the required identity.

Exam Tip: This is a variation of the standard three-variable identity \( x^3 + y^3 + z^3 = 3xyz \) when \( x + y + z = 0 \). Here, \( y \) is replaced by \( 2b \).

Question 8. Use the property: if \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \) to evaluate:
(i) \( 13^3 + (-8)^3 + (-5)^3 \)
(ii) \( 7^3 + 3^3 + (-10)^3 \)
(iii) \( 9^3 - 5^3 - 4^3 \)
(iv) \( 38^3 + (-26)^3 + (-12)^3 \)
Answer:
According to the algebraic property, if \( a + b + c = 0 \), then:
\( a^3 + b^3 + c^3 = 3abc \)
We apply this rule to each part:

(i) Let \( a = 13 \), \( b = -8 \), and \( c = -5 \).
Checking the sum: \( 13 + (-8) + (-5) = 13 - 13 = 0 \)
Therefore:
\( 13^3 + (-8)^3 + (-5)^3 = 3(13)(-8)(-5) = 1560 \)

(ii) Let \( a = 7 \), \( b = 3 \), and \( c = -10 \).
Checking the sum: \( 7 + 3 + (-10) = 10 - 10 = 0 \)
Therefore:
\( 7^3 + 3^3 + (-10)^3 = 3(7)(3)(-10) = -630 \)

(iii) We can rewrite \( 9^3 - 5^3 - 4^3 \) as \( 9^3 + (-5)^3 + (-4)^3 \).
Let \( a = 9 \), \( b = -5 \), and \( c = -4 \).
Checking the sum: \( 9 + (-5) + (-4) = 9 - 9 = 0 \)
Therefore:
\( 9^3 - 5^3 - 4^3 = 3(9)(-5)(-4) = 540 \)

(iv) Let \( a = 38 \), \( b = -26 \), and \( c = -12 \).
Checking the sum: \( 38 + (-26) + (-12) = 38 - 38 = 0 \)
Therefore:
\( 38^3 + (-26)^3 + (-12)^3 = 3(38)(-26)(-12) = 35568 \)
In simple words: If three numbers add up to zero, you can find the sum of their cubes instantly by multiplying the three numbers together and then multiplying by 3.

Exam Tip: Always show that the sum of the three bases is zero before applying this formula, as this condition is required for the identity to hold.

Question 9. If \( a - \frac{1}{a} = 3 \), find the value of:
(i) \( a^2 + \frac{1}{a^2} \)
(ii) \( a^3 - \frac{1}{a^3} \)
Answer:
Given:
\( a - \frac{1}{a} = 3 \)

(i) Squaring both sides of the given equation:
\( \left(a - \frac{1}{a}\right)^2 = 3^2 \)
\( \implies a^2 + \frac{1}{a^2} - 2 = 9 \)
\( \implies a^2 + \frac{1}{a^2} = 9 + 2 = 11 \)

(ii) Take the cube of both sides of \( a - \frac{1}{a} = 3 \):
\( \left(a - \frac{1}{a}\right)^3 = 3^3 \)
\( \implies a^3 - \frac{1}{a^3} - 3\left(a - \frac{1}{a}\right) = 27 \)
Substitute the value \( a - \frac{1}{a} = 3 \):
\( \implies a^3 - \frac{1}{a^3} - 3(3) = 27 \)
\( \implies a^3 - \frac{1}{a^3} - 9 = 27 \)
\( \implies a^3 - \frac{1}{a^3} = 27 + 9 = 36 \)
In simple words: First, square both sides to find the sum of squares. Second, cube both sides to find the difference of cubes.

Exam Tip: Remember that \( \left(a - \frac{1}{a}\right)^2 \) results in a \( -2 \) term, while \( \left(a - \frac{1}{a}\right)^3 \) results in a \( -3\left(a - \frac{1}{a}\right) \) term.

Question 10. If \( a - \frac{1}{a} = 4 \), find the value of:
(i) \( a^2 + \frac{1}{a^2} \)
(ii) \( a^4 + \frac{1}{a^4} \)
(iii) \( a^3 - \frac{1}{a^3} \)
Answer:
Given:
\( a - \frac{1}{a} = 4 \)

(i) We know that:
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \)
\( \implies a^2 + \frac{1}{a^2} = \left(a - \frac{1}{a}\right)^2 + 2 \)
Substitute the given value:
\( \implies a^2 + \frac{1}{a^2} = (4)^2 + 2 = 16 + 2 = 18 \) ....(1)

(ii) We know that:
\( a^4 + \frac{1}{a^4} = \left(a^2 + \frac{1}{a^2}\right)^2 - 2 \)
Substitute the value from (1):
\( \implies a^4 + \frac{1}{a^4} = (18)^2 - 2 = 324 - 2 = 322 \)

(iii) Using the identity:
\( \left(a - \frac{1}{a}\right)^3 = a^3 - \frac{1}{a^3} - 3\left(a - \frac{1}{a}\right) \)
\( \implies a^3 - \frac{1}{a^3} = \left(a - \frac{1}{a}\right)^3 + 3\left(a - \frac{1}{a}\right) \)
Substitute the given value:
\( \implies a^3 - \frac{1}{a^3} = (4)^3 + 3(4) = 64 + 12 = 76 \)
In simple words: We can build powers of 2, 3, and 4 sequentially by repeatedly squaring or cubing our starting terms.

Exam Tip: To get the fourth power, square the second power and subtract 2, since \( \left(x^2 + y^2\right)^2 = x^4 + y^4 + 2x^2y^2 \).

Question 11. If \( x + \frac{1}{x} = 2 \), show that \( x^2 + \frac{1}{x^2} = x^3 + \frac{1}{x^3} = x^4 + \frac{1}{x^4} \).
Answer:
We are given:
\( x + \frac{1}{x} = 2 \)
Let us evaluate each expression individually:

1. For \( x^2 + \frac{1}{x^2} \):
\( \left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \)
\( \implies x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \)
Substitute \( x + \frac{1}{x} = 2 \):
\( \implies x^2 + \frac{1}{x^2} = (2)^2 - 2 = 4 - 2 = 2 \) ....(1)

2. For \( x^3 + \frac{1}{x^3} \):
\( \left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) \)
\( \implies x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \)
Substitute \( x + \frac{1}{x} = 2 \):
\( \implies x^3 + \frac{1}{x^3} = (2)^3 - 3(2) = 8 - 6 = 2 \) ....(2)

3. For \( x^4 + \frac{1}{x^4} \):
\( x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2 \)
Substitute \( x^2 + \frac{1}{x^2} = 2 \) from equation (1):
\( \implies x^4 + \frac{1}{x^4} = (2)^2 - 2 = 4 - 2 = 2 \) ....(3)

From equations (1), (2), and (3), we get:
\( x^2 + \frac{1}{x^2} = x^3 + \frac{1}{x^3} = x^4 + \frac{1}{x^4} = 2 \)
Hence proved.
In simple words: When \( x + \frac{1}{x} = 2 \), any positive integer power of \( x \) added to its reciprocal will always result in 2.

Exam Tip: Label your steps with equation numbers (1), (2), and (3) to make the final comparison clear to the examiner.

Question 12. If \( 2x - 3y = 10 \) and \( xy = 16 \), find the value of \( 8x^3 - 27y^3 \).
Answer:
Given details:
\( 2x - 3y = 10 \) and \( xy = 16 \)
Taking the cube of both sides of the subtraction equation:
\( (2x - 3y)^3 = (10)^3 \)
\( \implies 8x^3 - 27y^3 - 3(2x)(3y)(2x - 3y) = 1000 \)
\( \implies 8x^3 - 27y^3 - 18xy(2x - 3y) = 1000 \)
Substitute the known values \( xy = 16 \) and \( 2x - 3y = 10 \):
\( \implies 8x^3 - 27y^3 - 18(16)(10) = 1000 \)
\( \implies 8x^3 - 27y^3 - 2880 = 1000 \)
\( \implies 8x^3 - 27y^3 = 1000 + 2880 \)
\( \implies 8x^3 - 27y^3 = 3880 \)
Thus, the value of the expression is 3880.
In simple words: Cube the expression \( 2x - 3y \), plug in the values for \( xy \) and \( 2x - 3y \), and simplify to solve for the target terms.

Exam Tip: Note that \( (2x)^3 = 8x^3 \) and \( (3y)^3 = 27y^3 \), which tells you that cubing the given linear equation is the correct strategy.

Question 13. Find the product of the following expressions:
(i) \( (3x + 5y + 2z)(3x - 5y + 2z) \)
(ii) \( (3x - 5y - 2z)(3x - 5y + 2z) \)
Answer:
(i) We group the terms to use the identity \( (a + b)(a - b) = a^2 - b^2 \):
\( (3x + 5y + 2z)(3x - 5y + 2z) = \{(3x + 2z) + 5y\}\{(3x + 2z) - 5y\} \)
\( \implies = (3x + 2z)^2 - (5y)^2 \)
Expanding the squared bracket:
\( \implies = 9x^2 + 4z^2 + 2 \times 3x \times 2z - 25y^2 \)
\( \implies = 9x^2 + 4z^2 + 12xz - 25y^2 \)
\( \implies = 9x^2 + 4z^2 - 25y^2 + 12xz \)

(ii) Grouping the terms similarly:
\( (3x - 5y - 2z)(3x - 5y + 2z) = \{(3x - 5y) - 2z\}\{(3x - 5y) + 2z\} \)
\( \implies = (3x - 5y)^2 - (2z)^2 \)
Expanding the square:
\( \implies = 9x^2 + 25y^2 - 2 \times 3x \times 5y - 4z^2 \)
\( \implies = 9x^2 + 25y^2 - 30xy - 4z^2 \)
\( \implies = 9x^2 + 25y^2 - 4z^2 - 30xy \)
In simple words: Group the terms that remain identical in both brackets together as 'a', and the term that changes its sign as 'b'. Then use the difference of squares formula.

Exam Tip: Grouping the terms correctly is the key to simplifying multi-variable expansions quickly.

Question 14. The sum of two numbers is 9 and their product is 20. Find:
(i) the sum of their squares
(ii) the sum of their cubes
Answer:
Let the two numbers be represented by \( a \) and \( b \).
As per the question:
\( a + b = 9 \) and \( ab = 20 \)

(i) For the sum of squares \( a^2 + b^2 \):
Square both sides of the sum equation:
\( (a + b)^2 = 9^2 \)
\( \implies a^2 + b^2 + 2ab = 81 \)
Substitute \( ab = 20 \):
\( \implies a^2 + b^2 + 2(20) = 81 \)
\( \implies a^2 + b^2 + 40 = 81 \)
\( \implies a^2 + b^2 = 81 - 40 = 41 \)
So, the sum of their squares is 41.

(ii) For the sum of cubes \( a^3 + b^3 \):
Cube both sides of the sum equation:
\( (a + b)^3 = 9^3 \)
\( \implies a^3 + b^3 + 3ab(a + b) = 729 \)
Substitute \( ab = 20 \) and \( a + b = 9 \):
\( \implies a^3 + b^3 + 3(20)(9) = 729 \)
\( \implies a^3 + b^3 + 540 = 729 \)
\( \implies a^3 + b^3 = 729 - 540 = 189 \)
So, the sum of their cubes is 189.
In simple words: Represent the numbers as variables, then square their sum to find the sum of squares, and cube their sum to find the sum of cubes.

Exam Tip: Writing down the starting equations \( a + b = 9 \) and \( ab = 20 \) earns initial marks, so always state your variables and equations clearly.

Question 15. If \( x - y = 5 \), \( x + y = 11 \), and \( xy = 24 \), find the values of:
(i) \( x^3 - y^3 \)
(ii) \( x^3 + y^3 \)
Answer:
We are given the following relations:
\( x - y = 5 \), \( x + y = 11 \), and \( xy = 24 \)

(i) To find the difference of cubes \( x^3 - y^3 \):
Take the cube of \( x - y = 5 \):
\( (x - y)^3 = 5^3 \)
\( \implies x^3 - y^3 - 3xy(x - y) = 125 \)
Substitute the given values \( xy = 24 \) and \( x - y = 5 \):
\( \implies x^3 - y^3 - 3(24)(5) = 125 \)
\( \implies x^3 - y^3 - 360 = 125 \)
\( \implies x^3 - y^3 = 125 + 360 = 485 \)
Hence, the difference of their cubes is 485.

(ii) To find the sum of cubes \( x^3 + y^3 \):
Take the cube of \( x + y = 11 \):
\( (x + y)^3 = 11^3 \)
\( \implies x^3 + y^3 + 3xy(x + y) = 1331 \)
Substitute the values \( xy = 24 \) and \( x + y = 11 \):
\( \implies x^3 + y^3 + 3(24)(11) = 1331 \)
\( \implies x^3 + y^3 + 792 = 1331 \)
\( \implies x^3 + y^3 = 1331 - 792 = 539 \)
Hence, the sum of their cubes is 539.
In simple words: Cube the subtraction equation to find the difference of the cubes, and cube the addition equation to find their sum.

Exam Tip: Double-check your arithmetic when cubing numbers like 11 to avoid calculation mistakes (\( 11^3 = 1331 \)).

Exercise 4(C)

Question 1. Expand the following products:
(i) \( (x + 8)(x + 10) \)
(ii) \( (x + 8)(x - 10) \)
(iii) \( (x - 8)(x + 10) \)
(iv) \( (x - 8)(x - 10) \)
Answer:
We use the algebraic identity:
\( (x + a)(x + b) = x^2 + (a + b)x + ab \)

(i) Here, \( a = 8 \) and \( b = 10 \):
\( (x + 8)(x + 10) = x^2 + (8 + 10)x + 8 \times 10 \)
\( \implies = x^2 + 18x + 80 \)

(ii) Here, \( a = 8 \) and \( b = -10 \):
\( (x + 8)(x - 10) = x^2 + (8 - 10)x + 8 \times (-10) \)
\( \implies = x^2 - 2x - 80 \)

(iii) Here, \( a = -8 \) and \( b = 10 \):
\( (x - 8)(x + 10) = x^2 + (-8 + 10)x + (-8) \times 10 \)
\( \implies = x^2 + 2x - 80 \)

(iv) Here, \( a = -8 \) and \( b = -10 \):
\( (x - 8)(x - 10) = x^2 + (-8 - 10)x + (-8)(-10) \)
\( \implies = x^2 - 18x + 80 \)
In simple words: Multiply the terms using the standard binomial identity, paying close attention to the positive or negative signs of the constant numbers.

Exam Tip: Note how changing the sign of the constant terms affects only the middle term's coefficient and the sign of the final constant.

Question 2. Multiply the following expressions:
(i) \( \left(2x - \frac{1}{x}\right)\left(3x + \frac{2}{x}\right) \)
(ii) \( \left(3a + \frac{2}{b}\right)\left(2a - \frac{3}{b}\right) \)
Answer:
(i) Expanding \( \left(2x - \frac{1}{x}\right)\left(3x + \frac{2}{x}\right) \) term-by-term:
\( \left(2x - \frac{1}{x}\right)\left(3x + \frac{2}{x}\right) = (2x)(3x) + (2x)\left(\frac{2}{x}\right) - \left(\frac{1}{x}\right)(3x) - \left(\frac{1}{x}\right)\left(\frac{2}{x}\right) \)
\( \implies = 6x^2 + 4 - 3 - \frac{2}{x^2} \)
\( \implies = 6x^2 + 1 - \frac{2}{x^2} \)

(ii) Expanding \( \left(3a + \frac{2}{b}\right)\left(2a - \frac{3}{b}\right) \) term-by-term:
\( \left(3a + \frac{2}{b}\right)\left(2a - \frac{3}{b}\right) = (3a)(2a) - (3a)\left(\frac{3}{b}\right) + \left(\frac{2}{b}\right)(2a) - \left(\frac{2}{b}\right)\left(\frac{3}{b}\right) \)
\( \implies = 6a^2 - \frac{9a}{b} + \frac{4a}{b} - \frac{6}{b^2} \)
\( \implies = 6a^2 + \left(\frac{4}{b} - \frac{9}{b}\right)a - \frac{6}{b^2} \)
\( \implies = 6a^2 - \frac{5a}{b} - \frac{6}{b^2} \)
In simple words: Multiply each term in the first bracket by each term in the second bracket, then combine any similar terms to simplify.

Exam Tip: When multiplying fractions like \( \frac{2}{b} \times 2a \), remember that the integer multiplies only the numerator.

Question 3. Expand the following squared trinomial expressions:
(i) \( (x + y - z)^2 \)
(ii) \( (x - 2y + 2)^2 \)
(iii) \( (5a - 3b + c)^2 \)
(iv) \( (5x - 3y - 2)^2 \)
(v) \( \left(x - \frac{1}{x} + 5\right)^2 \)
Answer:
We use the algebraic identity:
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \)

(i) Grouping terms for \( (x + y - z)^2 \):
\( (x + y - z)^2 = x^2 + y^2 + (-z)^2 + 2(x)(y) + 2(y)(-z) + 2(-z)(x) \)
\( \implies = x^2 + y^2 + z^2 + 2xy - 2yz - 2zx \)

(ii) Grouping terms for \( (x - 2y + 2)^2 \):
\( (x - 2y + 2)^2 = x^2 + (-2y)^2 + (2)^2 + 2(x)(-2y) + 2(-2y)(2) + 2(2)(x) \)
\( \implies = x^2 + 4y^2 + 4 - 4xy - 8y + 4x \)

(iii) Grouping terms for \( (5a - 3b + c)^2 \):
\( (5a - 3b + c)^2 = (5a)^2 + (-3b)^2 + c^2 + 2(5a)(-3b) + 2(-3b)(c) + 2(c)(5a) \)
\( \implies = 25a^2 + 9b^2 + c^2 - 30ab - 6bc + 10ca \)

(iv) Grouping terms for \( (5x - 3y - 2)^2 \):
\( (5x - 3y - 2)^2 = (5x)^2 + (-3y)^2 + (-2)^2 + 2(5x)(-3y) + 2(-3y)(-2) + 2(-2)(5x) \)
\( \implies = 25x^2 + 9y^2 + 4 - 30xy + 12y - 20x \)

(v) Grouping terms for \( \left(x - \frac{1}{x} + 5\right)^2 \):
\( \left(x - \frac{1}{x} + 5\right)^2 = (x)^2 + \left(-\frac{1}{x}\right)^2 + (5)^2 + 2(x)\left(-\frac{1}{x}\right) + 2\left(-\frac{1}{x}\right)(5) + 2(5)(x) \)
\( \implies = x^2 + \frac{1}{x^2} + 25 - 2 - \frac{10}{x} + 10x \)
\( \implies = x^2 + \frac{1}{x^2} + 23 - \frac{10}{x} + 10x \)
In simple words: Treat each term inside the bracket as a separate variable (with its sign attached) and apply the trinomial squaring formula.

Exam Tip: Pay attention to sign rules: the square of any negative term always becomes positive, but the product terms can be negative.

Question 4. If \( a^2 + b^2 + c^2 = 50 \) and \( a + b + c = 12 \), find the value of \( ab + bc + ca \).
Answer:
We know the identity:
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \) ....(1)
Given that \( a^2 + b^2 + c^2 = 50 \) and \( a + b + c = 12 \).
We substitute these values into the identity (1):
\( (12)^2 = 50 + 2(ab + bc + ca) \)
\( \implies 144 = 50 + 2(ab + bc + ca) \)
Subtracting 50 from both sides:
\( \implies 94 = 2(ab + bc + ca) \)
\( \implies ab + bc + ca = \frac{94}{2} \)
\( \implies ab + bc + ca = 47 \)
Thus, the value of \( ab + bc + ca \) is 47.
In simple words: We can find the sum of products of two variables at a time by substituting the known values of the sum and sum of squares into the trinomial expansion formula.

Exam Tip: Rearrange the equation cleanly before performing the final division to ensure you don't make any arithmetic mistakes.

Question 5. If \( a^2 + b^2 + c^2 = 35 \) and \( ab + bc + ca = 23 \), find the value of \( a + b + c \).
Answer:
We write down the standard trinomial identity:
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \) ....(1)
Given that:
\( a^2 + b^2 + c^2 = 35 \) and \( ab + bc + ca = 23 \)
Substitute these given values into (1):
\( (a + b + c)^2 = 35 + 2(23) \)
\( \implies (a + b + c)^2 = 35 + 46 \)
\( \implies (a + b + c)^2 = 81 \)
Taking the square root on both sides:
\( \implies a + b + c = \pm\sqrt{81} \)
\( \implies a + b + c = \pm 9 \)
Hence, the value is \( \pm 9 \).
In simple words: Substitute the given sum of squares and product pairs into the identity, sum them up, and then find the square root.

Exam Tip: Don't forget that square roots have both positive and negative values. Always write \( \pm 9 \) to score full marks.

Question 6. If \( a + b + c = p \) and \( ab + bc + ca = q \), find the value of \( a^2 + b^2 + c^2 \).
Answer:
We use the algebraic relation:
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \) ....(1)
Given values:
\( a + b + c = p \) and \( ab + bc + ca = q \)
Substitute these variables into equation (1):
\( (p)^2 = a^2 + b^2 + c^2 + 2(q) \)
\( \implies p^2 = a^2 + b^2 + c^2 + 2q \)
To isolate our target term, rearrange the equation:
\( \implies a^2 + b^2 + c^2 = p^2 - 2q \)
Therefore, the value of \( a^2 + b^2 + c^2 \) is \( p^2 - 2q \).
In simple words: Insert the letters \( p \) and \( q \) into the trinomial identity and rearrange the terms to solve for the sum of squares.

Exam Tip: Expressing the answer purely in terms of the given parameters (\( p \) and \( q \)) is the objective of this problem.

Question 7. If \( a^2 + b^2 + c^2 = 50 \) and \( ab + bc + ca = 47 \), find the value of \( a + b + c \).
Answer:
We know that:
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
Given information:
\( a^2 + b^2 + c^2 = 50 \) and \( ab + bc + ca = 47 \)
Substituting these values:
\( (a + b + c)^2 = 50 + 2(47) \)
\( \implies (a + b + c)^2 = 50 + 94 \)
\( \implies (a + b + c)^2 = 144 \)
Taking square roots on both sides:
\( \implies a + b + c = \pm\sqrt{144} \)
\( \implies a + b + c = \pm 12 \)
So, the required value of \( a + b + c \) is \( \pm 12 \).
In simple words: Add twice the sum of product terms to the sum of squares, then find the square root of the final total.

Exam Tip: Ensure you multiply \( 47 \) by \( 2 \) before adding it to \( 50 \) to respect the order of operations.

Question 8. If \( x + y - z = 4 \) and \( x^2 + y^2 + z^2 = 30 \), find the value of \( xy - yz - zx \).
Answer:
Given:
\( x + y - z = 4 \) and \( x^2 + y^2 + z^2 = 30 \)
Consider the expansion of \( (x + y - z)^2 \):
\( (x + y - z)^2 = x^2 + y^2 + z^2 + 2(xy - yz - zx) \)
Substitute the given values into this equation:
\( (4)^2 = 30 + 2(xy - yz - zx) \)
\( \implies 16 = 30 + 2(xy - yz - zx) \)
Rearranging the equation to solve for the group of terms:
\( \implies 2(xy - yz - zx) = 16 - 30 \)
\( \implies 2(xy - yz - zx) = -14 \)
Dividing both sides by 2:
\( \implies xy - yz - zx = \frac{-14}{2} \)
\( \implies xy - yz - zx = -7 \)
Hence, the required value is \( -7 \).
In simple words: Square the linear equation with its minus sign intact, plug in the sum of squares, and solve for the combined product terms.

Exam Tip: Notice how the minus sign of \( z \) changes the signs of the product terms containing \( z \) into negative signs.

Exercise 4(D)

Question 1. If \( x + 2y + 3z = 0 \) and \( x^3 + 4y^3 + 9z^3 = 18xyz \), find the value of:
\( \frac{(x + 2y)^2}{xy} + \frac{(2y + 3z)^2}{yz} + \frac{(3z + x)^2}{zx} \)
Answer:
We are given:
\( x + 2y + 3z = 0 \) and \( x^3 + 4y^3 + 9z^3 = 18xyz \)
From the first equation, we can write:
\( x + 2y = -3z \)
\( 2y + 3z = -x \)
\( 3z + x = -2y \)
Now substitute these into the target expression:
\( \frac{(x + 2y)^2}{xy} + \frac{(2y + 3z)^2}{yz} + \frac{(3z + x)^2}{zx} \)
\( \implies = \frac{(-3z)^2}{xy} + \frac{(-x)^2}{yz} + \frac{(-2y)^2}{zx} \)
\( \implies = \frac{9z^2}{xy} + \frac{x^2}{yz} + \frac{4y^2}{zx} \)
To add these fractions, find a common denominator \( xyz \):
\( \implies = \frac{9z^3 + x^3 + 4y^3}{xyz} \)
\( \implies = \frac{x^3 + 4y^3 + 9z^3}{xyz} \)
Since we are given \( x^3 + 4y^3 + 9z^3 = 18xyz \), we substitute this in:
\( \implies = \frac{18xyz}{xyz} = 18 \)
Thus, the value of the expression is 18.
In simple words: Use the sum equation to substitute for the bracketed terms, find a common denominator to combine them, and then substitute the cubic value.

Exam Tip: Isolating parts of the linear equation (like \( x + 2y = -3z \)) is the easiest way to simplify complex fractions.

Question 2. If \( a + \frac{1}{a} = m \), find the values of:
(i) \( a - \frac{1}{a} \)
(ii) \( a^2 - \frac{1}{a^2} \)
Answer:
Given:
\( a + \frac{1}{a} = m \)

(i) Let us find the square of the given relation:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)
\( \implies m^2 = a^2 + \frac{1}{a^2} + 2 \)
\( \implies a^2 + \frac{1}{a^2} = m^2 - 2 \)
Now, consider the expansion of \( \left(a - \frac{1}{a}\right)^2 \):
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \)
Substitute \( a^2 + \frac{1}{a^2} = m^2 - 2 \) into this equation:
\( \implies \left(a - \frac{1}{a}\right)^2 = (m^2 - 2) - 2 \)
\( \implies \left(a - \frac{1}{a}\right)^2 = m^2 - 4 \)
Taking the square root:
\( \implies a - \frac{1}{a} = \pm\sqrt{m^2 - 4} \) ....(1)

(ii) Using the difference of squares identity:
\( a^2 - \frac{1}{a^2} = \left(a + \frac{1}{a}\right)\left(a - \frac{1}{a}\right) \)
Substitute the known values \( a + \frac{1}{a} = m \) and \( a - \frac{1}{a} = \pm\sqrt{m^2 - 4} \):
\( \implies a^2 - \frac{1}{a^2} = m \left(\pm\sqrt{m^2 - 4}\right) \)
\( \implies a^2 - \frac{1}{a^2} = \pm m\sqrt{m^2 - 4} \)
In simple words: First, relate the square of the difference to the square of the sum to find \( a - \frac{1}{a} \). Then, multiply both expressions together to find the difference of their squares.

Exam Tip: Remember that \( \left(a + \frac{1}{a}\right)^2 - \left(a - \frac{1}{a}\right)^2 = 4 \). This direct relationship is a useful shortcut for such problems.

Question 3. In the expansion of \( \left(2x^2 - 8\right)(x - 4)^2 \), find:
(i) the coefficient of \( x^3 \)
(ii) the coefficient of \( x^2 \)
(iii) the constant term
Answer:
Let us expand the expression \( \left(2x^2 - 8\right)(x - 4)^2 \):
First, expand the squared binomial:
\( (x - 4)^2 = x^2 - 8x + 16 \)
Now multiply the two terms:
\( \left(2x^2 - 8\right)\left(x^2 - 8x + 16\right) = 2x^2\left(x^2 - 8x + 16\right) - 8\left(x^2 - 8x + 16\right) \)
\( \implies = 2x^4 - 16x^3 + 32x^2 - 8x^2 + 64x - 128 \)
Combine the like terms \( 32x^2 \) and \( -8x^2 \):
\( \implies = 2x^4 - 16x^3 + 24x^2 + 64x - 128 \)
From this fully expanded expression:
- The coefficient of \( x^3 \) is \( -16 \)
- The coefficient of \( x^2 \) is \( 24 \)
- The constant term is \( -128 \)
In simple words: Fully expand both brackets by multiplying every term out, collect the similar terms together, and read off the numbers next to the required powers of \( x \).

Exam Tip: Be careful with the signs when distributing \( -8 \) across the terms of the second bracket.

Question 4. If \( x^2 + \frac{1}{9x^2} = \frac{25}{36} \), find the value of \( x + \frac{1}{3x} \).
Answer:
We are given that:
\( x^2 + \frac{1}{9x^2} = \frac{25}{36} \)
\( \implies x^2 + \frac{1}{(3x)^2} = \frac{25}{36} \) ....(1)

Let us expand the expression \( \left(x + \frac{1}{3x}\right)^2 \):
\( \left(x + \frac{1}{3x}\right)^2 = x^2 + \frac{1}{(3x)^2} + 2 \times x \times \frac{1}{3x} \)
\( \implies \left(x + \frac{1}{3x}\right)^2 = x^2 + \frac{1}{(3x)^2} + \frac{2}{3} \)
By substituting equation (1) in this expansion, we get:
\( \implies \left(x + \frac{1}{3x}\right)^2 = \frac{25}{36} + \frac{2}{3} \)
\( \implies \left(x + \frac{1}{3x}\right)^2 = \frac{25 + 24}{36} \)
\( \implies \left(x + \frac{1}{3x}\right)^2 = \frac{49}{36} \)
Taking the square root on both sides:
\( \implies x + \frac{1}{3x} = \pm \sqrt{\frac{49}{36}} \)
\( \implies x + \frac{1}{3x} = \pm \frac{7}{6} \) ....(2)
In simple words: To find the value, we square the target expression. Then we replace the known part with its given value, simplify the fraction, and take the square root.

Exam Tip: Always remember to write the plus-minus sign (\( \pm \)) when taking a square root, as both positive and negative values are mathematically valid here.

Question 5. If \( 2(x^2 + 1) = 5x \), find the value of:
(i) \( x - \frac{1}{x} \)
(ii) \( x^3 - \frac{1}{x^3} \)
Answer:
(i) We start with the given equation:
\( 2(x^2 + 1) = 5x \)
\( \implies x^2 + 1 = \frac{5}{2}x \)
Divide both sides by \( x \) to get:
\( \frac{x^2+1}{x} = \frac{5}{2} \)
\( \implies x + \frac{1}{x} = \frac{5}{2} \) ....(1)

Next, expand the squared term \( \left(x + \frac{1}{x}\right)^2 \):
\( \left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \)
Substitute equation (1) into this:
\( \implies \left(\frac{5}{2}\right)^2 = x^2 + \frac{1}{x^2} + 2 \)
\( \implies \left(\frac{5}{2}\right)^2 - 2 = x^2 + \frac{1}{x^2} \)
\( \implies \frac{25}{4} - 2 = x^2 + \frac{1}{x^2} \)
\( \implies x^2 + \frac{1}{x^2} = \frac{25-8}{4} \)
\( \implies x^2 + \frac{1}{x^2} = \frac{17}{4} \) ....(2)

Now, let us expand \( \left(x - \frac{1}{x}\right)^2 \):
\( \left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \)
\( \implies \left(x - \frac{1}{x}\right)^2 = \frac{17}{4} - 2 \)
\( \implies \left(x - \frac{1}{x}\right)^2 = \frac{17-8}{4} \)
\( \implies \left(x - \frac{1}{x}\right)^2 = \frac{9}{4} \)
Taking square roots on both sides:
\( \implies x - \frac{1}{x} = \pm \frac{3}{2} \) ....(3)

(ii) Using the algebraic identity:
\( x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right) \)
Now, substitute the value from equation (3) into this identity:
\( \implies x^3 - \frac{1}{x^3} = \left(\pm \frac{3}{2}\right)^3 + 3\left(\pm \frac{3}{2}\right) \)
\( = \pm \frac{27}{8} \pm \frac{9}{2} \)
\( \implies x^3 - \frac{1}{x^3} = \pm \frac{27+36}{8} \)
\( \implies x^3 - \frac{1}{x^3} = \pm \frac{63}{8} \)
In simple words: First, divide by x to find \( x + 1/x \). Then find \( x^2 + 1/x^2 \) using squaring, use that to get \( x - 1/x \), and finally apply the cubic formula to solve the second part.

Exam Tip: Be careful with the signs during the cubic expansion; when you cube \( \pm \), the output carries the same \( \pm \) sign.

Question 6. If \( a^2 + b^2 = 34 \) and \( ab = 12 \), find the value of:
(i) \( 3(a+b)^2 + 5(a-b)^2 \)
(ii) \( 7(a-b)^2 - 2(a+b)^2 \)
Answer:
We can find the values of \( (a+b)^2 \) and \( (a-b)^2 \) first:
\( (a+b)^2 = a^2 + b^2 + 2ab \)
Substitute the given values:
\( = 34 + 2(12) \)
\( = 34 + 24 \)
\( = 58 \)

Similarly, for \( (a-b)^2 \):
\( (a-b)^2 = a^2 + b^2 - 2ab \)
\( = 34 - 2(12) \)
\( = 34 - 24 \)
\( = 10 \)

Now we can evaluate the given expressions:
(i) \( 3(a+b)^2 + 5(a-b)^2 \)
\( = 3(58) + 5(10) \)
\( = 174 + 50 \)
\( = 224 \)

(ii) \( 7(a-b)^2 - 2(a+b)^2 \)
\( = 7(10) - 2(58) \)
\( = 70 - 116 \)
\( = -46 \)
In simple words: First calculate the values of the squared sum and the squared difference by inserting the given numbers into standard algebraic expansions. Then plug those results into the two final expressions.

Exam Tip: Work out the values of \( (a+b)^2 \) and \( (a-b)^2 \) as separate steps to simplify the main calculations and avoid silly arithmetic errors.

Question 7. If \( 3x - \frac{4}{x} = 4 \), find the value of \( 27x^3 - \frac{64}{x^3} \).
Answer:
Let us cube both sides by expanding \( \left(3x - \frac{4}{x}\right)^3 \):
\( \left(3x - \frac{4}{x}\right)^3 = (3x)^3 - \left(\frac{4}{x}\right)^3 - 3(3x)\left(\frac{4}{x}\right)\left(3x - \frac{4}{x}\right) \)
\( \implies \left(3x - \frac{4}{x}\right)^3 = 27x^3 - \frac{64}{x^3} - 36\left(3x - \frac{4}{x}\right) \)
Substitute the given value \( 3x - \frac{4}{x} = 4 \) into this equation:
\( \implies (4)^3 = 27x^3 - \frac{64}{x^3} - 36(4) \)
\( \implies 64 = 27x^3 - \frac{64}{x^3} - 144 \)
Rearranging the equation:
\( \implies 27x^3 - \frac{64}{x^3} = 64 + 144 \)
\( \implies 27x^3 - \frac{64}{x^3} = 208 \)
In simple words: Cube both sides of the given equation. This lets you use the identity for a difference of cubes, substitute the known value, and solve for the target term.

Exam Tip: When using the identity \( (a-b)^3 = a^3 - b^3 - 3ab(a-b) \), ensure the middle term simplifies cleanly by cancelling out the \( x \) variables.

Question 8. If \( x^2 + \frac{1}{x^2} = 7 \), find the value of \( 7x^3 + 8x - \frac{7}{x^3} - \frac{8}{x} \).
Answer:
Starting with the equation we are given:
\( x^2 + \frac{1}{x^2} = 7 \)
Subtract 2 from both sides of the equation:
\( \implies x^2 + \frac{1}{x^2} - 2 = 7 - 2 \)
\( \implies \left(x - \frac{1}{x}\right)^2 = 5 \)
\( \implies x - \frac{1}{x} = \sqrt{5} \) ....(1)

Now, let us group the terms of the expression we need to find:
\( 7x^3 + 8x - \frac{7}{x^3} - \frac{8}{x} = 7x^3 - \frac{7}{x^3} + 8x - \frac{8}{x} \)
\( = 7\left(x^3 - \frac{1}{x^3}\right) + 8\left(x - \frac{1}{x}\right) \) ....(2)

Next, expand \( \left(x - \frac{1}{x}\right)^3 \):
\( \left(x - \frac{1}{x}\right)^3 = x^3 - \frac{1}{x^3} - 3\left(x - \frac{1}{x}\right) \)
\( \implies x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right) \)
Substitute the value from equation (1):
\( \implies x^3 - \frac{1}{x^3} = (\sqrt{5})^3 + 3(\sqrt{5}) \) ....(3)

Now substitute equation (3) and equation (1) back into equation (2):
\( 7x^3 + 8x - \frac{7}{x^3} - \frac{8}{x} = 7\left[(\sqrt{5})^3 + 3\sqrt{5}\right] + 8\left[\sqrt{5}\right] \)
\( \implies 7x^3 + 8x - \frac{7}{x^3} - \frac{8}{x} = 7[5\sqrt{5} + 3\sqrt{5}] + 8\sqrt{5} \)
\( \implies 7x^3 + 8x - \frac{7}{x^3} - \frac{8}{x} = 7[8\sqrt{5}] + 8\sqrt{5} \)
\( \implies 7x^3 + 8x - \frac{7}{x^3} - \frac{8}{x} = 56\sqrt{5} + 8\sqrt{5} = 64\sqrt{5} \)
In simple words: Find \( x - 1/x \) first. Then regroup the target expression into two parts: a cubic part and a linear part, find their values, and combine them.

Exam Tip: Grouping terms with common coefficients (like 7 and 8 here) makes it much easier to substitute known algebraic identities.

Question 9. If \( x = \frac{1}{x-5} \), find the value of \( x^2 - \frac{1}{x^2} \).
Answer:
By cross-multiplying the given equation:
\( x(x-5) = 1 \)
\( \implies x^2 - 5x = 1 \)
\( \implies x^2 - 1 = 5x \)
Divide each term by \( x \):
\( \frac{x^2-1}{x} = 5 \)
\( \implies x - \frac{1}{x} = 5 \) ....(1)

Now, let us square both sides of this equation:
\( \left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \)
\( \implies (5)^2 = x^2 + \frac{1}{x^2} - 2 \)
\( \implies x^2 + \frac{1}{x^2} = 25 + 2 = 27 \) ....(2)

Next, consider the expansion of \( \left(x + \frac{1}{x}\right)^2 \):
\( \left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \)
\( \implies \left(x + \frac{1}{x}\right)^2 = 27 + 2 \)
\( \implies \left(x + \frac{1}{x}\right)^2 = 29 \)
Taking square roots on both sides:
\( \implies x + \frac{1}{x} = \pm \sqrt{29} \) ....(3)

We can write the difference of squares as:
\( x^2 - \frac{1}{x^2} = \left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right) \)
Substitute the values from equations (1) and (3):
\( = (\pm \sqrt{29})(5) \)
\( \implies x^2 - \frac{1}{x^2} = \pm 5\sqrt{29} \)
In simple words: Convert the starting equation into a simpler form to find \( x - 1/x \). Use this to find \( x^2 + 1/x^2 \), then calculate \( x + 1/x \), and multiply both to find the difference of their squares.

Exam Tip: Remember the standard algebraic factorization \( a^2 - b^2 = (a-b)(a+b) \) to evaluate differences of squares.

Question 10. If \( x = \frac{1}{5-x} \), find the value of \( x^3 + \frac{1}{x^3} \).
Answer:
By cross-multiplying the given expression:
\( x(5-x) = 1 \)
\( \implies 5x - x^2 = 1 \)
Rearranging the terms:
\( \implies x^2 + 1 = 5x \)
Divide each term by \( x \):
\( \frac{x^2+1}{x} = 5 \)
\( \implies x + \frac{1}{x} = 5 \) ....(1)

Now, we use the standard algebraic identity for sum of cubes:
\( x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \)
Substitute the value from equation (1):
\( = (5)^3 - 3(5) \)
\( \implies x^3 + \frac{1}{x^3} = 125 - 15 = 110 \)
In simple words: Cross-multiply to get an equation in terms of \( x^2 \) and \( x \). Dividing by \( x \) gives \( x + 1/x \), which can be plugged directly into the cubic identity.

Exam Tip: Be careful with the minus sign in the identity \( a^3+b^3 = (a+b)^3 - 3ab(a+b) \); a common mistake is using a plus sign instead of minus.

Question 11. If \( 3a + 5b + 4c = 0 \), show that \( 27a^3 + 125b^3 + 64c^3 = 180abc \).
Answer:
We are given:
\( 3a + 5b + 4c = 0 \)
We can rewrite this as:
\( 3a + 5b = -4c \)

Now, cube both sides of the equation:
\( (3a + 5b)^3 = (-4c)^3 \)
\( \implies (3a)^3 + (5b)^3 + 3(3a)(5b)(3a + 5b) = -64c^3 \)
\( \implies 27a^3 + 125b^3 + 45ab(3a + 5b) = -64c^3 \)
Substitute \( 3a + 5b = -4c \) back into the equation:
\( \implies 27a^3 + 125b^3 + 45ab(-4c) = -64c^3 \)
\( \implies 27a^3 + 125b^3 - 180abc = -64c^3 \)
Rearranging the terms:
\( \implies 27a^3 + 125b^3 + 64c^3 = 180abc \)
Hence proved.
In simple words: Shift one term to the right side and cube both sides. Replace that shifted term back into the expansion, and simplify to get the desired result.

Exam Tip: This can also be proven using the identity \( x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \). Since \( x+y+z = 0 \), the right side becomes 0, yielding \( x^3+y^3+z^3 = 3xyz \).

Question 12. The sum of two numbers is 7 and the sum of their cubes is 133. Find the sum of their squares.
Answer:
Let the two numbers be \( a \) and \( b \).
Based on the problem statement:
\( a + b = 7 \)
\( a^3 + b^3 = 133 \)

Using the identity for the cube of a sum:
\( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \)
\( \implies (7)^3 = 133 + 3ab(7) \)
\( \implies 343 = 133 + 21ab \)
\( \implies 21ab = 343 - 133 \)
\( \implies 21ab = 210 \)
\( \implies ab = 10 \)

Next, we calculate the sum of their squares using the identity:
\( a^2 + b^2 = (a+b)^2 - 2ab \)
Substitute the known values:
\( = 7^2 - 2(10) \)
\( = 49 - 20 \)
\( = 29 \)
In simple words: Represent the numbers as a and b. Use the cubic expansion formula to find their product (ab), and then use the squared expansion formula to find the sum of their squares.

Exam Tip: Finding the product \( ab \) is an intermediate but crucial step; always look for ways to relate the given sums of higher powers to the basic product.

Question 13. Find the value of \( a \) if:
(i) \( 4x^2 + ax + 9 = (2x+3)^2 \)
(ii) \( 4x^2 + ax + 9 = (2x-3)^2 \)
(iii) \( 9x^2 + (7a-5)x + 25 = (3x+5)^2 \)
Answer:
(i) We start by expanding the right side of the equation:
\( (2x + 3)^2 = 4x^2 + 12x + 9 \)
Now, compare the coefficients of \( x \) on both sides:
\( ax = 12x \)
\( \implies a = 12 \)

(ii) Expand the right side of the second equation:
\( (2x - 3)^2 = 4x^2 - 12x + 9 \)
Comparing the \( x \) coefficients on both sides:
\( ax = -12x \)
\( \implies a = -12 \)

(iii) Expand the right side of the third equation:
\( (3x + 5)^2 = 9x^2 + 30x + 25 \)
Comparing the coefficient of \( x \) terms on both sides:
\( 7a - 5 = 30 \)
\( \implies 7a = 35 \)
\( \implies a = 5 \)
In simple words: Expand the bracket on the right side using squaring identities, and then find 'a' by matching the middle terms.

Exam Tip: Be extremely careful with signs when equating coefficients; a negative sign in front of a term completely changes the final value of the variable.

Question 14. If \( \frac{x^2+1}{x} = \frac{10}{3} \), find the value of \( x^3 - \frac{1}{x^3} \).
Answer:
The given equation is:
\( \frac{x^2+1}{x} = \frac{10}{3} \)
\( \implies x + \frac{1}{x} = \frac{10}{3} \)

Square both sides to find \( x^2 + \frac{1}{x^2} \):
\( \implies x^2 + \frac{1}{x^2} + 2 = \frac{100}{9} \)
\( \implies x^2 + \frac{1}{x^2} = \frac{100}{9} - 2 = \frac{82}{9} \)

Now, we calculate \( x - \frac{1}{x} \):
\( \implies x - \frac{1}{x} = \sqrt{\left(x + \frac{1}{x}\right)^2 - 4} \)
\( = \sqrt{\frac{100}{9} - 4} \)
\( = \sqrt{\frac{64}{9}} = \frac{8}{3} \)

Cube both sides of this result:
\( \left(x - \frac{1}{x}\right)^3 = \left(\frac{8}{3}\right)^3 \)
\( \implies x^3 - \frac{1}{x^3} - 3\left(x - \frac{1}{x}\right) = \frac{512}{27} \)
Substitute \( x - \frac{1}{x} = \frac{8}{3} \) into this:
\( \implies x^3 - \frac{1}{x^3} - 3\left(\frac{8}{3}\right) = \frac{512}{27} \)
\( \implies x^3 - \frac{1}{x^3} - 8 = \frac{512}{27} \)
\( \implies x^3 - \frac{1}{x^3} = \frac{512}{27} + 8 \)
\( \implies x^3 - \frac{1}{x^3} = \frac{512 + 216}{27} = \frac{728}{27} \)
In simple words: Simplify the given equation to find \( x + 1/x \). Use that to get \( x - 1/x \), and then cube both sides to find the difference of their cubes.

Exam Tip: Relate the sum and difference of reciprocal terms using the identity \( (x-1/x)^2 = (x+1/x)^2 - 4 \) to transition between them smoothly.

Question 15. The difference between two positive numbers is 4 and the difference between their cubes is 316. Find the sum of their squares.
Answer:
Let the two positive numbers be represented by \( a \) and \( b \).
According to the given information:
\( a - b = 4 \)
\( a^3 - b^3 = 316 \)

Cube both sides of the first equation:
\( (a - b)^3 = 4^3 \)
\( \implies a^3 - b^3 - 3ab(a-b) = 64 \)
Substitute \( a^3 - b^3 = 316 \) and \( a - b = 4 \):
\( \implies 316 - 3ab(4) = 64 \)
\( \implies 316 - 64 = 12ab \)
\( \implies 252 = 12ab \)
\( \implies ab = 21 \)

Next, square both sides of the first equation:
\( (a-b)^2 = 16 \)
\( \implies a^2 + b^2 - 2ab = 16 \)
Substitute \( ab = 21 \) into this relation:
\( \implies a^2 + b^2 - 2(21) = 16 \)
\( \implies a^2 + b^2 = 16 + 42 = 58 \)
In simple words: Call the numbers a and b. Use the cubic difference to find their product (ab), and then use the squared difference to find the sum of their squares.

Exam Tip: Be sure to write down the steps of cubing and squaring the linear equation clearly, as intermediate steps like finding \( ab \) carry specific marks.

Exercise 4(E)

Question 1. Use the identity \( (x+a)(x+b)(x+c) = x^3 + (a+b+c)x^2 + (ab+bc+ca)x + abc \) to expand the following:
(i) \( (x+6)(x+4)(x-2) \)
(ii) \( (x-6)(x-4)(x+2) \)
(iii) \( (x-6)(x-4)(x-2) \)
(iv) \( (x+6)(x-4)(x-2) \)
Answer:
Using the identity, we can expand each expression:

(i) Here, \( a=6 \), \( b=4 \), and \( c=-2 \):
\( (x+6)(x+4)(x-2) = x^3 + (6+4-2)x^2 + [6 \times 4 + 4 \times (-2) + (-2) \times 6]x + 6 \times 4 \times (-2) \)
\( = x^3 + 8x^2 + (24 - 8 - 12)x - 48 \)
\( = x^3 + 8x^2 + 4x - 48 \)

(ii) Here, \( a=-6 \), \( b=-4 \), and \( c=2 \):
\( (x-6)(x-4)(x+2) = x^3 + (-6-4+2)x^2 + [(-6) \times (-4) + (-4) \times 2 + 2 \times (-6)]x + (-6) \times (-4) \times 2 \)
\( = x^3 - 8x^2 + (24 - 8 - 12)x + 48 \)
\( = x^3 - 8x^2 + 4x + 48 \)

(iii) Here, \( a=-6 \), \( b=-4 \), and \( c=-2 \):
\( (x-6)(x-4)(x-2) = x^3 + (-6-4-2)x^2 + [(-6) \times (-4) + (-4) \times (-2) + (-2) \times (-6)]x + (-6) \times (-4) \times (-2) \)
\( = x^3 - 12x^2 + (24 + 8 + 12)x - 48 \)
\( = x^3 - 12x^2 + 44x - 48 \)

(iv) Here, \( a=6 \), \( b=-4 \), and \( c=-2 \):
\( (x+6)(x-4)(x-2) = x^3 + (6-4-2)x^2 + [6 \times (-4) + (-4) \times (-2) + (-2) \times 6]x + 6 \times (-4) \times (-2) \)
\( = x^3 + 0x^2 + (-24 + 8 - 12)x + 48 \)
\( = x^3 - 28x + 48 \)
In simple words: Identify the three values for a, b, and c including their signs. Substitute them into the formula and calculate the coefficient for each power of x.

Exam Tip: Pay close attention to negative signs during the multiplication of pairs; a single sign error will make the final coefficient incorrect.

Question 2. Simplify each of the following expressions:
(i) \( (2x+3y)(4x^2 - 6xy + 9y^2) \)
(ii) \( \left(3x - \frac{5}{x}\right)\left(9x^2 + 15 + \frac{25}{x^2}\right) \)
(iii) \( \left(\frac{a}{3} - 3b\right)\left(\frac{a^2}{9} + ab + 9b^2\right) \)
Answer:
We can simplify these products using the sum/difference of cubes identities:
\( (A+B)(A^2 - AB + B^2) = A^3 + B^3 \)
\( (A-B)(A^2 + AB + B^2) = A^3 - B^3 \)

(i) For \( (2x+3y)(4x^2 - 6xy + 9y^2) \):
This matches the identity with \( A = 2x \) and \( B = 3y \):
\( = (2x+3y)[(2x)^2 - (2x)(3y) + (3y)^2] \)
\( = (2x)^3 + (3y)^3 \)
\( = 8x^3 + 27y^3 \)

(ii) For \( \left(3x - \frac{5}{x}\right)\left(9x^2 + 15 + \frac{25}{x^2}\right) \):
This matches the identity with \( A = 3x \) and \( B = \frac{5}{x} \):
\( = \left(3x - \frac{5}{x}\right)\left[(3x)^2 + (3x)\left(\frac{5}{x}\right) + \left(\frac{5}{x}\right)^2\right] \)
\( = (3x)^3 - \left(\frac{5}{x}\right)^3 \)
\( = 27x^3 - \frac{125}{x^3} \)

(iii) For \( \left(\frac{a}{3} - 3b\right)\left(\frac{a^2}{9} + ab + 9b^2\right) \):
This matches the identity with \( A = \frac{a}{3} \) and \( B = 3b \):
\( = \left(\frac{a}{3} - 3b\right)\left[\left(\frac{a}{3}\right)^2 + \left(\frac{a}{3}\right)(3b) + (3b)^2\right] \)
\( = \left(\frac{a}{3}\right)^3 - (3b)^3 \)
\( = \frac{a^3}{27} - 27b^3 \)
In simple words: Rewrite each bracket to fit the standard sum or difference of cubes formula. Once you match the terms, simply cube the first and second terms.

Exam Tip: Always verify that the middle term of the trinomial matches \( AB \) before applying the sum or difference of cubes formula directly.

Question 3. Find the cube of each of the following using algebraic identities:
(i) \( 104 \)
(ii) \( 97 \)
Answer:
(i) We can write \( 104 \) as \( (100 + 4) \):
Using the identity \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \):
\( (104)^3 = (100 + 4)^3 \)
\( = (100)^3 + (4)^3 + 3(100)(4)(100 + 4) \)
\( = 1000000 + 64 + 1200(104) \)
\( = 1000000 + 64 + 124800 \)
\( = 1124864 \)

(ii) We can write \( 97 \) as \( (100 - 3) \):
Using the identity \( (a-b)^3 = a^3 - b^3 - 3ab(a-b) \):
\( (97)^3 = (100 - 3)^3 \)
\( = (100)^3 - (3)^3 - 3(100)(3)(100 - 3) \)
\( = 1000000 - 27 - 900(97) \)
\( = 1000000 - 27 - 87300 \)
\( = 912673 \)
In simple words: Rewrite the number as a sum or difference of 100 and a small number, then apply the cubic expansion formula to calculate easily without actual long multiplication.

Exam Tip: Splitting numbers using base 100 makes calculations very straightforward. Double-check your subtraction steps when using the negative identity.

Question 4. Simplify: \( \frac{(x^2-y^2)^3 + (y^2-z^2)^3 + (z^2-x^2)^3}{(x-y)^3 + (y-z)^3 + (z-x)^3} \).
Answer:
Recall the algebraic property: if \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \).

First, let us examine the numerator terms:
Let \( a = x^2 - y^2 \), \( b = y^2 - z^2 \), and \( c = z^2 - x^2 \).
Since \( a + b + c = (x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2) = 0 \), we have:
\( \implies (x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3 = 3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2) \) ....(1)

Next, examine the denominator terms:
Let \( A = x - y \), \( B = y - z \), and \( C = z - x \).
Since \( A + B + C = (x - y) + (y - z) + (z - x) = 0 \), we have:
\( \implies (x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x) \) ....(2)

Substituting (1) and (2) back into the given expression:
\( \frac{3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)}{3(x - y)(y - z)(z - x)} \)
\( = \frac{(x-y)(x+y)(y-z)(y+z)(z-x)(z+x)}{(x-y)(y-z)(z-x)} \)
Canceling the matching factors from numerator and denominator:
\( = (x+y)(y+z)(z+x) \)
In simple words: Since the sum of the inner terms in both the top and bottom expressions is zero, we can replace their sum of cubes with three times their product, factor them, and cancel terms.

Exam Tip: Recognizing the conditional identity \( a^3+b^3+c^3 = 3abc \) saves complex expansion work and is a very common topic in exams.

Question 5. Evaluate using algebraic identities:
(i) \( \frac{0.8 \times 0.8 \times 0.8 + 0.5 \times 0.5 \times 0.5}{0.8 \times 0.8 - 0.8 \times 0.5 + 0.5 \times 0.5} \)
(ii) \( \frac{1.2 \times 1.2 + 1.2 \times 0.3 + 0.3 \times 0.3}{1.2 \times 1.2 \times 1.2 - 0.3 \times 0.3 \times 0.3} \)
Answer:
(i) Let \( a = 0.8 \) and \( b = 0.5 \).
The expression can be written as:
\( \frac{a \times a \times a + b \times b \times b}{a \times a - a \times b + b \times b} = \frac{a^3 + b^3}{a^2 - ab + b^2} \)
Using the identity \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \):
\( = \frac{(a+b)(a^2 - ab + b^2)}{a^2 - ab + b^2} \)
\( = a + b \)
Substitute the actual values back:
\( = 0.8 + 0.5 = 1.3 \)

(ii) Let \( a = 1.2 \) and \( b = 0.3 \).
The expression can be written as:
\( \frac{a \times a + a \times b + b \times b}{a \times a \times a - b \times b \times b} = \frac{a^2 + ab + b^2}{a^3 - b^3} \)
Using the identity \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \):
\( = \frac{a^2 + ab + b^2}{(a-b)(a^2 + ab + b^2)} \)
\( = \frac{1}{a-b} \)
Substitute the values back:
\( = \frac{1}{1.2 - 0.3} = \frac{1}{0.9} = \frac{10}{9} = 1\frac{1}{9} \)
In simple words: Substitute variables for the decimals to view the algebraic layout. Apply the cubic factor identity to cancel matching terms and then do a quick final calculation.

Exam Tip: Never compute decimal multiplications directly in such questions; rewriting the terms using standard algebraic identities is much quicker and avoids mistakes.

Question 6. If \( a - 2b + 3c = 0 \), find the value of \( a^3 - 8b^3 + 27c^3 \).
Answer:
We can write the given expression as a sum of cubes:
\( a^3 - 8b^3 + 27c^3 = a^3 + (-2b)^3 + (3c)^3 \)

We are given that:
\( a - 2b + 3c = 0 \)
\( \implies a + (-2b) + 3c = 0 \)

Recall that if the sum of three terms is zero (\( x + y + z = 0 \)), then the sum of their cubes equals three times their product (\( x^3 + y^3 + z^3 = 3xyz \)).
Applying this rule here:
\( a^3 + (-2b)^3 + (3c)^3 = 3(a)(-2b)(3c) \)
\( = -18abc \)
In simple words: Rewrite the cubic expression as a sum of cubed terms. Since the sum of these base numbers is zero, their cubed sum equals three times their product.

Exam Tip: Ensure you include the negative sign of \( -2b \) inside the bracket when setting up the sum so that the final product carries the correct negative sign.

Question 7. If \( x + 5y = 10 \), show that \( x^3 + 125y^3 + 150xy - 1000 = 0 \).
Answer:
We start with the equation:
\( x + 5y = 10 \)

Now, cube both sides of the equation:
\( (x + 5y)^3 = 10^3 \)
Using the identity \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \):
\( \implies x^3 + (5y)^3 + 3(x)(5y)(x + 5y) = 1000 \)
\( \implies x^3 + 125y^3 + 15xy(x + 5y) = 1000 \)
Substitute the given value of \( x + 5y = 10 \) back into this equation:
\( \implies x^3 + 125y^3 + 15xy(10) = 1000 \)
\( \implies x^3 + 125y^3 + 150xy = 1000 \)
Subtract 1000 from both sides to get the desired form:
\( \implies x^3 + 125y^3 + 150xy - 1000 = 0 \)
In simple words: Cube both sides of the equation, use the standard algebraic identity, substitute the known linear sum back, and move all terms to one side.

Exam Tip: Remember to substitute the original value \( x+5y = 10 \) into the intermediate product step to get the linear term \( 150xy \).

Question 8. If \( x = 3 + 2\sqrt{2} \), find the value of:
(i) \( \frac{1}{x} \)
(ii) \( x - \frac{1}{x} \)
(iii) \( \left(x - \frac{1}{x}\right)^3 \)
(iv) \( x^3 - \frac{1}{x^3} \)
Answer:
(i) To find \( \frac{1}{x} \), we rationalize the denominator:
\( \frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} \)
Multiply the numerator and denominator by the conjugate \( 3 - 2\sqrt{2} \):
\( = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \)
\( = \frac{3 - 2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2} \)
\( = \frac{3 - 2\sqrt{2}}{9 - 8} \)
\( \implies \frac{1}{x} = 3 - 2\sqrt{2} \) ....(1)

(ii) Now, calculate \( x - \frac{1}{x} \):
\( x - \frac{1}{x} = (3 + 2\sqrt{2}) - (3 - 2\sqrt{2}) \)
\( = 3 + 2\sqrt{2} - 3 + 2\sqrt{2} \)
\( \implies x - \frac{1}{x} = 4\sqrt{2} \) ....(2)

(iii) Find the value of \( \left(x - \frac{1}{x}\right)^3 \):
Using the result from equation (2):
\( \left(x - \frac{1}{x}\right)^3 = (4\sqrt{2})^3 \)
\( = 64 \times 2\sqrt{2} \)
\( = 128\sqrt{2} \)

(iv) Lastly, find \( x^3 - \frac{1}{x^3} \):
Using the identity \( x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right) \):
\( = 128\sqrt{2} + 3(4\sqrt{2}) \)
\( = 128\sqrt{2} + 12\sqrt{2} \)
\( = 140\sqrt{2} \)
In simple words: First rationalise the reciprocal of x. Subtract this from x to find the linear difference, cube that difference for part three, and apply the cubic identity to find part four.

Exam Tip: Rationalizing the denominator carefully by multiplying by its conjugate is the key to simplifying surd-based expressions.

Question 9. If \( a + b = 11 \) and \( a^2 + b^2 = 65 \), find the value of \( a^3 + b^3 \).
Answer:
First, find the product \( ab \) by expanding \( (a+b)^2 \):
\( (a+b)^2 = a^2 + b^2 + 2ab \)
Substitute our known values:
\( \implies (11)^2 = 65 + 2ab \)
\( \implies 121 = 65 + 2ab \)
\( \implies 2ab = 121 - 65 \)
\( \implies 2ab = 56 \)
\( \implies ab = 28 \)

Now, we evaluate \( a^3 + b^3 \) using the identity:
\( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
Substitute the values into this formula:
\( = (11)(65 - 28) \)
\( = 11 \times 37 \)
\( = 407 \)
In simple words: Square the sum to find the product of the two numbers. Then use the sum of cubes factorization identity to compute the final answer.

Exam Tip: Writing down both forms of the cubic identity, including \( a^3+b^3 = (a+b)^3 - 3ab(a+b) \), is helpful; either version will lead to the same result.