GSEB Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.5

Question 1. Use suitable identities to find the following products:
(i) \( (x + 4)(x + 10) \)
(ii) \( (x + 8)(x - 10) \)
(iii) \( (3x + 4)(3x - 5) \)
(iv) \( (y^2 + \frac {3}{2})(y^2 - \frac {3}{2}) \)
(v) \( (3 - 2x)(3 + 2x) \)
Answer:
(i) Using the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \), we obtain:
\( (x + 4)(x + 10) = x^2 + (4 + 10)x + 4 \times 10 \)
\( = x^2 + 14x + 40 \)
In simple words: We use a special algebra rule to multiply these two expressions. It helps us find the answer quickly by putting in the numbers for 'a' and 'b'.
(ii) Using the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \), we obtain:
\( (x + 8)(x - 10) \)
\( = (x + 8){x + (-10)} \)
\( = x^2 + {8 + (-10)}x + 8 \times (-10) \)
\( = x^2 - 2x - 80 \)
In simple words: Applying the same algebraic rule, we substitute the values, remembering that -10 acts like 'b'. This gives us a simplified polynomial.
(iii) Using the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \), we obtain:
\( (3x + 4)(3x - 5) \)
\( = (3x)^2 + {4 + (-5)}(3x) + 4 \times (-5) \)
\( = 9x^2 + (-1)3x - 20 \)
\( = 9x^2 - 3x - 20 \)
In simple words: For this one, 3x acts like 'x' in our formula. We substitute carefully, and the negative numbers are handled correctly to get the final answer.
(iv) Using the identity \( (a + b)(a - b) = a^2 - b^2 \), we obtain:
\( (y^2 + \frac {3}{2})(y^2 - \frac {3}{2}) = (y^2)^2 - (\frac {3}{2})^2 \)
\( = y^4 - \frac {9}{4} \)
In simple words: This uses a different rule for multiplication, where if you have two terms that are the same but one has a plus and the other a minus, you just square each term and subtract.
(v) Using the identity \( (a - b)(a + b) = a^2 - b^2 \), we obtain:
\( (3 - 2x)(3 + 2x) = 3^2 - (2x)^2 \)
\( = 9 - 4x^2 \)
In simple words: We apply the same rule as before. Here, 3 is 'a' and 2x is 'b'. Square them both and subtract to get the simple form.

Exam Tip: Always identify the correct algebraic identity before attempting to solve. Careful substitution of values for a, b, and x is key to preventing mistakes.

Question 2. Evaluate the following products without multiplying directly:
(i) \( 103 \times 107 \)
(ii) \( 95 \times 96 \)
(iii) \( 104 \times 96 \)
Answer:
(i) We can write \( 103 \times 107 = (100 + 3)(100 + 7) \).
Using the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \), we get:
\( (100 + 3)(100 + 7) \)
\( = 100^2 + (3 + 7) \times 100 + 3 \times 7 \)
\( = 10000 + 10 \times 100 + 21 \)
\( = 10000 + 1000 + 21 \)
\( = 11021 \)
In simple words: We break down the numbers into easier parts like 100 plus something. Then we use an algebra rule to multiply them quickly without doing a long calculation.
(ii) We can write \( 95 \times 96 = (100 - 5)(100 - 4) \).
Using the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \), we get:
\( {100 + (-5)}{100 + (-4)} \)
\( = 100^2 + {(-5) + (-4)} \times 100 + (-5) \times (-4) \)
\( = 10000 + (-9) \times 100 + 20 \)
\( = 10000 - 900 + 20 \)
\( = 9120 \)
In simple words: We see that 95 and 96 are close to 100, so we write them as 100 minus a number. Then we apply the same algebra rule, being careful with the negative signs when we add and multiply.
(iii) We can write \( 104 \times 96 = (100 + 4)(100 - 4) \).
Using the identity \( (a + b)(a - b) = a^2 - b^2 \), we get:
\( (100 + 4)(100 - 4) = (100)^2 - 4^2 \)
\( = 10000 - 16 \)
\( = 9984 \)
In simple words: This one is easy because the numbers are like (a+b) and (a-b). So we just square the first part, square the second part, and subtract them.

Exam Tip: For multiplication problems, look for numbers close to powers of 10 (100, 1000) or simple factors to apply identities effectively.

Question 3. Factorise the following using appropriate identities:
(i) \( 9x^2 + 6xy + y^2 \)
(ii) \( 4y^2 - 4y + 1 \)
(iii) \( x^2 - \frac{y^{2}}{100} \)
Answer:
(i) We need to factorise \( 9x^2 + 6xy + y^2 \).
We can rewrite it as \( (3x)^2 + 2 \times 3x \times y + (y)^2 \).
This matches the identity \( a^2 + 2ab + b^2 = (a + b)^2 \).
Therefore, \( 9x^2 + 6xy + y^2 = (3x + y)^2 \).
In simple words: We look for a pattern that matches a known algebra rule. Here, it looks like "something squared + two times something times something else + something else squared". Once we find that pattern, we can write it as (first thing + second thing) squared.
(ii) We need to factorise \( 4y^2 - 4y + 1 \).
We can rewrite it as \( (2y)^2 - 2 \times 2y \times 1 + 1^2 \).
This matches the identity \( a^2 - 2ab + b^2 = (a - b)^2 \).
Therefore, \( 4y^2 - 4y + 1 = (2y - 1)^2 \).
In simple words: This is very similar to the last one, but with a minus sign in the middle. So it follows the rule for (first thing - second thing) squared. We just identify the 'a' and 'b' parts correctly.
(iii) We need to factorise \( x^2 - \frac{y^{2}}{100} \).
We can rewrite it as \( x^2 - (\frac{y}{10})^2 \).
This matches the identity \( a^2 - b^2 = (a - b)(a + b) \).
Therefore, \( x^2 - \frac{y^{2}}{100} = (x - \frac{y}{10})(x + \frac{y}{10}) \).
In simple words: When we see one squared term minus another squared term, we use the "difference of squares" rule. This rule says we can write it as (first thing minus second thing) times (first thing plus second thing).

Exam Tip: Recognising the square terms (like \( 9x^2 = (3x)^2 \)) and the middle term \( (2ab) \) is essential for quickly applying perfect square identities.

Question 4. Expand each of the following using suitable identities:
(i) \( (x + 2y +4z)^2 \)
(ii) \( (2x - y + z)^2 \)
(iii) \( (-2x + 3y + 2z)^2 \)
(iv) \( (3a - 7b - c)^2 \)
(v) \( (-2x + 5y - 3z)^2 \)
(vi) \( [\frac {1}{4} a – \frac {1}{2} b + 1]^2 \)
Answer:
(i) Using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \), we expand:
\( (x + 2y + 4z)^2 \)
\( = (x)^2 + (2y)^2 + (4z)^2 + 2 \times x \times 2y + 2 \times 2y \times 4z + 2 \times 4z \times x \)
\( = x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx \)
In simple words: This is a rule for squaring three terms added together. You square each term, then add twice the product of each pair of terms.
(ii) Using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \), we expand:
\( (2x - y + z)^2 \)
\( = {2x + (-y) + z}^2 \)
\( = (2x)^2 + (-y)^2 + z^2 + 2 \times 2x \times (-y) + 2 \times (-y) \times z + 2 \times z \times 2x \)
\( = 4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx \)
In simple words: We apply the same rule, but we treat the 'minus y' as 'plus negative y' to fit the formula. This ensures all signs are handled correctly.
(iii) Using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \), we expand:
\( (-2x + 3y + 2z)^2 \)
\( = (-2x)^2 + (3y)^2 + (2z)^2 + 2 \times (-2x) \times 3y + 2 \times 3y \times 2z + 2 \times 2z \times (-2x) \)
\( = 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx \)
In simple words: Again, we follow the same pattern, being very careful with the negative sign on the -2x term throughout the calculation. Squaring a negative gives a positive.
(iv) Using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \), we expand:
\( (3a - 7b - c)^2 \)
\( = {3a + (-7b) + (-c)}^2 \)
\( = (3a)^2 + (-7b)^2 + (-c)^2 + 2 \times 3a \times (-7b) + 2 \times (-7b) \times (-c) + 2 \times (-c) \times 3a \)
\( = 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca \)
In simple words: Here we have multiple negative terms. We imagine them as additions of negative numbers, then apply the three-term squaring rule, which helps in managing all the signs.
(v) Using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \), we expand:
\( (-2x + 5y - 3z)^2 \)
\( = {(-2x) + 5y + (-3z)}^2 \)
\( = (-2x)^2 + (5y)^2 + (-3z)^2 + 2 \times (-2x) \times 5y + 2 \times 5y \times (-3z) + 2 \times (-3z) \times (-2x) \)
\( = 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12zx \)
In simple words: This is another example of expanding three terms, where two of them are negative. We keep converting subtractions to additions of negative numbers and apply the same formula.
(vi) Using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \), we expand:
\( [\frac {1}{4} a – \frac {1}{2} b + 1]^2 \)
\( = [\frac {1}{4} a + (-\frac {1}{2}b) + 1]^2 \)
\( = (\frac {1}{4} a)^2 + (-\frac {1}{2} b)^2 + 1^2 + 2 \times \frac {1}{4} a \times (-\frac {1}{2}b) + 2 \times (-\frac {1}{2}b) \times 1 + 2 \times 1 \times \frac {1}{4} a \)
\( = \frac{a^{2}}{16} + \frac{b^{2}}{4} + 1 - \frac {1}{4}ab - b + \frac {1}{2}a \)
In simple words: Even with fractions, the rule remains the same. We square each part and then add twice the product of each possible pair, carefully managing the fractions and negative signs.

Exam Tip: Be extra careful with negative signs when substituting into the identity \( (a+b+c)^2 \). A common mistake is to forget to include the negative sign with 'b' or 'c' in the \( 2ab \), \( 2bc \), \( 2ca \) terms.

Question 5. Factorise:
(i) \( 4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz \)
(ii) \( 2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4 \sqrt{2}yz - 8zx \)
Answer:
(i) We need to factorise \( 4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz \).
We look for terms that can be squared: \( (2x)^2 \), \( (3y)^2 \), \( (4z)^2 \).
Since \( -24yz \) and \( -16xz \) are negative, one of the variables (x, y, or z) must have a negative coefficient. Since the \( xy \) term is positive \( (12xy) \), x and y must have the same sign. Thus, z must be negative.
So, we can write it as \( (2x)^2 + (3y)^2 + (-4z)^2 + 2 \times (2x) \times (3y) + 2 \times (3y) \times (-4z) + 2 \times (-4z) \times (2x) \).
This matches the identity \( a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2 \).
Therefore, \( 4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz = (2x + 3y - 4z)^2 \).
Which can also be written as \( (2x + 3y - 4z)(2x + 3y - 4z) \).
In simple words: This expression looks like a squared sum of three terms. We find the square roots of the squared parts (like 2x, 3y, 4z). Then, we check the signs of the middle terms to figure out which of these square roots should be negative in the final answer. We see that 'z' must be negative because the 'yz' and 'xz' terms are negative.
(ii) We need to factorise \( 2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8zx \).
We can rewrite the terms as squares: \( (\sqrt{2}x)^2 \), \( (y)^2 \), \( (2\sqrt{2}z)^2 \).
From the term \( -2\sqrt{2}xy \), either \( \sqrt{2}x \) or \( y \) must be negative. Given the terms \( 4\sqrt{2}yz \) (positive) and \( -8zx \) (negative), we deduce that \( \sqrt{2}x \) must be negative.
So, we rewrite it as \( (-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 + 2 \times (-\sqrt{2}x) \times y + 2 \times y \times (2\sqrt{2}z) + 2 \times (2\sqrt{2}z) \times (-\sqrt{2}x) \).
This matches the identity \( a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (a + b + c)^2 \).
Therefore, \( 2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8zx = (-\sqrt{2}x + y + 2\sqrt{2}z)^2 \).
Which can also be written as \( (-\sqrt{2}x + y + 2\sqrt{2}z)(-\sqrt{2}x + y + 2\sqrt{2}z) \).
In simple words: This is another complex factorisation problem. We identify the parts that are squared, even if they have square roots. By looking at the signs of the mixed terms (like xy, yz, zx), we figure out which individual part should be negative. Then we apply the rule for squaring three terms.

Exam Tip: For factorisation problems involving squares and multiple variables, determine the signs of the square root terms by carefully examining the signs of the cross-product terms (e.g., \( 2xy \), \( 2yz \), \( 2zx \)).

Question 6. Write the following cubes in expanded form:
(i) \( (2x + 1)^3 \)
(ii) \( (2a - 3b)^3 \)
(iii) \( (\frac{3}{2} x + 1)^3 \)
(iv) \( (x - \frac{2}{3} y)^3 \)
Answer:
(i) Using the identity \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \), we expand:
\( (2x + 1)^3 \)
\( = (2x)^3 + 1^3 + 3 \times 2x \times 1(2x + 1) \)
\( = 8x^3 + 1 + 6x(2x + 1) \)
\( = 8x^3 + 1 + 12x^2 + 6x \)
\( = 8x^3 + 12x^2 + 6x + 1 \)
In simple words: We use the rule for cubing two terms added together. You cube each term, then add three times their product multiplied by their sum.
(ii) Using the identity \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \), we expand:
\( (2a - 3b)^3 \)
\( = (2a)^3 - (3b)^3 - 3 \times 2a \times 3b(2a - 3b) \)
\( = 8a^3 - 27b^3 - 18ab(2a - 3b) \)
\( = 8a^3 - 27b^3 - 36a^2b + 54ab^2 \)
In simple words: This is similar, but for two terms being subtracted and then cubed. The signs change, but the structure of cubing each term and involving three times their product remains.
(iii) Using the identity \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \), we expand:
\( (\frac{3}{2} x + 1)^3 \)
\( = (\frac{3}{2}x)^3 + 1^3 + 3 \times \frac{3}{2}x \times 1 (\frac{3}{2}x + 1) \)
\( = \frac{27}{8} x^3 + 1 + \frac{9}{2}x (\frac{3}{2}x + 1) \)
\( = \frac{27}{8} x^3 + 1 + \frac{27}{4} x^2 + \frac{9}{2}x \)
\( = \frac{27}{8} x^3 + \frac{27}{4} x^2 + \frac{9}{2}x + 1 \)
In simple words: Even with fractions, the cubing rule stays the same. We carefully cube the fractional term and multiply out the other parts, remembering to combine like terms.
(iv) Using the identity \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \), we expand:
\( (x - \frac{2}{3} y)^3 \)
\( = x^3 - (\frac{2}{3} y)^3 - 3 \times x \times \frac{2}{3} y (x - \frac{2}{3}y) \)
\( = x^3 - \frac{8}{27}y^3 - 2xy(x - \frac{2}{3} y) \)
\( = x^3 - \frac{8}{27}y^3 - 2x^2y + \frac{4 x y^{2}}{3} \)
\( = x^3 - 2x^2y + \frac{4 x y^{2}}{3} - \frac{8}{27} y^3 \)
In simple words: We apply the subtraction cubing rule here. This means cubing the first and second terms, then subtracting three times their product multiplied by their difference, ensuring fraction calculations are correct.

Exam Tip: Remember to cube both the coefficient and the variable when expanding terms like \( (2x)^3 \). Distribute the \( 3ab \) term correctly inside the parenthesis \( (a \pm b) \).

Question 7. Evaluate the following using suitable identities:
(i) \( (99)^3 \)
(ii) \( (102)^3 \)
(iii) \( (998)^3 \)
Answer:
(i) We can write \( (99)^3 = (100 - 1)^3 \).
Using the identity \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \), we get:
\( (100 - 1)^3 \)
\( = 100^3 - 1^3 - 3 \times 100 \times 1(100 - 1) \)
\( = 1000000 - 1 - 300(99) \)
\( = 1000000 - 1 - 29700 \)
\( = 970299 \)
In simple words: We change 99 to (100-1) so we can use a cubing rule easily. Then we calculate each part, subtracting the terms carefully to get the final answer.
(ii) We can write \( (102)^3 = (100 + 2)^3 \).
Using the identity \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \), we get:
\( (100 + 2)^3 \)
\( = 100^3 + 2^3 + 3 \times 100 \times 2(100 + 2) \)
\( = 1000000 + 8 + 600(102) \)
\( = 1000000 + 8 + 61200 \)
\( = 1061208 \)
In simple words: We change 102 to (100+2) to use the sum cubing rule. We cube each part, multiply and add them up, being careful with the numbers to get the large result.
(iii) We can write \( (998)^3 = (1000 - 2)^3 \).
Using the identity \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \), we get:
\( (1000 - 2)^3 \)
\( = 1000^3 - 2^3 - 3 \times 1000 \times 2(1000 - 2) \)
\( = 1000000000 - 8 - 6000(998) \)
\( = 1000000000 - 8 - 5988000 \)
\( = 994011992 \)
In simple words: We rewrite 998 as (1000-2). Then we use the difference of cubes rule. We cube 1000 and 2, then subtract three times their product multiplied by their difference, resulting in a very large number.

Exam Tip: Always choose the closest power of 10 to represent the number (e.g., 99 as 100-1, 102 as 100+2) to simplify calculations significantly.

Question 8. Factorise each of the following:
(i) \( 8a^3 + b^3 + 12a^2b + 6ab^2 \)
(ii) \( 8a^3 - b^3 - 12a^2b + 6ab^2 \)
(iii) \( 27 - 125a^3 - 135a + 225a^2 \)
(iv) \( 64a^3 – 27b^3 – 144a^2b + 108ab^2 \)
(v) \( 27p^3 - \frac {1}{216} – \frac {9}{2} p^2 + \frac {1}{4}p \)
Answer:
(i) We need to factorise \( 8a^3 + b^3 + 12a^2b + 6ab^2 \).
We can rewrite it as \( (2a)^3 + b^3 + 3 \times (2a)^2 \times b + 3 \times (2a) \times b^2 \).
Alternatively, \( (2a)^3 + b^3 + 3 \times 2a \times b(2a + b) \).
This matches the identity \( a^3 + b^3 + 3a^2b + 3ab^2 = (a + b)^3 \).
Therefore, \( 8a^3 + b^3 + 12a^2b + 6ab^2 = (2a + b)^3 \).
This can also be written as \( (2a + b)(2a + b)(2a + b) \).
In simple words: We look for the pattern of a cubed sum. We find the cube roots of the first two terms (2a and b), then check if the other terms fit the "3a squared b" and "3ab squared" pattern. If they do, it's (a+b) cubed.
(ii) We need to factorise \( 8a^3 - b^3 - 12a^2b + 6ab^2 \).
We can rewrite it as \( (2a)^3 - b^3 - 3 \times (2a)^2 \times b + 3 \times (2a) \times b^2 \).
Alternatively, \( (2a)^3 - b^3 - 3 \times 2a \times b (2a - b) \).
This matches the identity \( a^3 - b^3 - 3a^2b + 3ab^2 = (a - b)^3 \).
Therefore, \( 8a^3 - b^3 - 12a^2b + 6ab^2 = (2a - b)^3 \).
This can also be written as \( (2a - b)(2a - b)(2a - b) \).
In simple words: This looks like a cubed difference. We find the cube roots (2a and b). Then we check if the signs match the "3a squared b" and "3ab squared" terms for a difference cube. If they do, it's (a-b) cubed.
(iii) We need to factorise \( 27 - 125a^3 - 135a + 225a^2 \).
We can rewrite it as \( 3^3 - (5a)^3 - 3 \times 3 \times 5a (3 - 5a) \).
This matches the identity \( a^3 - b^3 - 3ab(a - b) = (a - b)^3 \).
Therefore, \( 27 - 125a^3 - 135a + 225a^2 = (3 - 5a)^3 \).
This can also be written as \( (3 - 5a)(3 - 5a)(3 - 5a) \).
In simple words: We rearrange the terms to match the pattern of a difference cube. We identify 3 as 'a' and 5a as 'b'. Once the pattern fits, we write it as (3 minus 5a) cubed.
(iv) We need to factorise \( 64a^3 - 27b^3 - 144a^2b + 108ab^2 \).
We can rewrite it as \( (4a)^3 - (3b)^3 - 3 \times 4a \times 3b(4a - 3b) \).
This matches the identity \( a^3 - b^3 - 3ab(a - b) = (a - b)^3 \).
Therefore, \( 64a^3 - 27b^3 - 144a^2b + 108ab^2 = (4a - 3b)^3 \).
This can also be written as \( (4a - 3b)(4a - 3b)(4a - 3b) \).
In simple words: This is another difference cube problem. We find the cube roots (4a and 3b) and confirm the middle terms match the pattern of minus 3ab(a-b). Then we write it as (4a minus 3b) cubed.
(v) We need to factorise \( 27p^3 - \frac {1}{216} – \frac {9}{2} p^2 + \frac {1}{4}p \).
We can rewrite it as \( (3p)^3 - (\frac {1}{6})^3 - 3 \times 3p \times \frac {1}{6} (3p - \frac {1}{6}) \).
This matches the identity \( a^3 - b^3 - 3ab(a - b) = (a - b)^3 \).
Therefore, \( 27p^3 - \frac {1}{216} – \frac {9}{2} p^2 + \frac {1}{4}p = (3p - \frac {1}{6})^3 \).
This can also be written as \( (3p - \frac {1}{6})(3p - \frac {1}{6})(3p - \frac {1}{6}) \).
In simple words: We identify the terms that are cubed, even with fractions. Then we check if the remaining terms fit the \( -3ab(a-b) \) part of the difference of cubes identity. If it fits, we write the factored form.

Exam Tip: For factorisation, be systematic. First, identify if it's a sum/difference of cubes. Then, find the 'a' and 'b' terms and verify that the intermediate terms \( (3a^2b \pm 3ab^2) \) match the expression.

Question 9. Verify:
(i) \( x^3 + y^3 = (x + y)(x^2 – xy + y^2) \)
(ii) \( x^3 – y^3 = (x - y)(x^2 + xy + y^2) \)
Answer:
(i) We know that \( (x + y)^3 = x^3 + y^3 + 3xy (x + y) \).
We want to show \( x^3 + y^3 \). So, we can rearrange the identity:
\( x^3 + y^3 = (x + y)^3 - 3xy(x + y) \)
Now, we can take \( (x + y) \) as a common factor:
\( x^3 + y^3 = (x + y){(x + y)^2 - 3xy} \)
Expand \( (x + y)^2 \):
\( x^3 + y^3 = (x + y){x^2 + y^2 + 2xy - 3xy} \)
Simplify the terms inside the curly brackets:
\( x^3 + y^3 = (x + y){x^2 + y^2 - xy} \)
Hence, \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \). Verified.
In simple words: To prove this, we start with the known rule for \( (x+y)^3 \). We then move the \( 3xy(x+y) \) part to the other side to get \( x^3+y^3 \) alone. After that, we factor out \( (x+y) \) and simplify the remaining part to match the formula we wanted to verify.
(ii) We know that \( (x - y)^3 = x^3 - y^3 - 3xy(x - y) \).
We want to show \( x^3 - y^3 \). So, we can rearrange the identity:
\( x^3 - y^3 = (x - y)^3 + 3xy(x - y) \)
Now, we can take \( (x - y) \) as a common factor:
\( x^3 - y^3 = (x - y){(x - y)^2 + 3xy} \)
Expand \( (x - y)^2 \):
\( x^3 - y^3 = (x - y){x^2 + y^2 - 2xy + 3xy} \)
Simplify the terms inside the curly brackets:
\( x^3 - y^3 = (x - y){x^2 + y^2 + xy} \)
Hence, \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \). Verified.
In simple words: To verify this one, we start with the formula for \( (x-y)^3 \). We isolate \( x^3-y^3 \) by moving the \( -3xy(x-y) \) term. Then we factor out \( (x-y) \) and simplify the remaining terms inside the brackets to show it matches the desired expression.

Exam Tip: When verifying identities, start with a known expansion (like \( (x+y)^3 \) or \( (x-y)^3 \)) and manipulate it algebraically to arrive at the target identity. Show each step clearly.

Question 10. Factorise each of the following:
(i) \( 27y^3 + 125z^3 \)
(ii) \( 64m^3 - 343n^3 \)
Answer:
(i) We need to factorise \( 27y^3 + 125z^3 \).
We can rewrite this as \( (3y)^3 + (5z)^3 \).
Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), we get:
\( (3y)^3 + (5z)^3 = (3y + 5z){(3y)^2 - (3y)(5z) + (5z)^2} \)
\( = (3y + 5z)(9y^2 - 15yz + 25z^2) \)
In simple words: We notice this is a sum of two cubes. So we use the rule that lets us factor it into two parts: a sum of the cube roots, and then a longer part that includes squares and a product.
(ii) We need to factorise \( 64m^3 - 343n^3 \).
We can rewrite this as \( (4m)^3 - (7n)^3 \).
Using the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \), we get:
\( (4m)^3 - (7n)^3 = (4m - 7n){(4m)^2 + (4m)(7n) + (7n)^2} \)
\( = (4m - 7n)(16m^2 + 28mn + 49n^2) \)
In simple words: This is a difference of two cubes. We use the specific rule for this, which gives us two factors: a difference of the cube roots, and a second factor with squared terms and a positive product term.

Exam Tip: Memorise the factorisation formulas for sum and difference of cubes: \( a^3 + b^3 \) and \( a^3 - b^3 \). This allows for direct application once 'a' and 'b' are identified.

Question 11. Factorise: \( 27x^3 + y^3 + z^3 - 9xyz \)
Answer:
We need to factorise \( 27x^3 + y^3 + z^3 - 9xyz \).
We can rewrite this as \( (3x)^3 + y^3 + z^3 - 3 \times (3x) \times y \times z \).
This matches the identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \).
Here, \( a = 3x \), \( b = y \), and \( c = z \).
Therefore, \( 27x^3 + y^3 + z^3 - 9xyz \)
\( = (3x + y + z){(3x)^2 + y^2 + z^2 - (3x)y - yz - z(3x)} \)
\( = (3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx) \)
In simple words: This expression has three cubed terms minus three times their product. This specific form means we can factor it into two main parts: the sum of the three terms, and a second part that includes their squares and subtracted products of pairs.

Exam Tip: The identity \( a^3 + b^3 + c^3 - 3abc \) is crucial here. Recognise the structure where three terms are cubed and their product is multiplied by 3 and subtracted.

Question 12. Verify that \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)((x - y)^2 + (y - z)^2 + (z - x)^2) \)
Answer:
We know the identity: \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \).
Now, let's work on the right side of the expression we need to verify:
\( (x + y + z) \frac{1}{2} [2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx] \)
We multiplied and divided by 2 to create terms that can be grouped into squares.
\( = (x + y + z) \frac{1}{2} [(x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2)] \)
Now, group the terms to form perfect squares:
\( = \frac{1}{2} (x + y + z) [(x - y)^2 + (y - z)^2 + (z - x)^2] \)
This matches the expression we needed to verify. Thus, the identity is proven.
In simple words: We start with a known rule for \( x^3+y^3+z^3-3xyz \). Then, we take the right side of the expression we need to prove and multiply it by \( \frac{1}{2} \times 2 \). This trick allows us to rearrange the terms inside the brackets into three perfect square differences, like \( (x-y)^2 \), which then matches the rule.

Exam Tip: To verify this particular identity, remember the trick of multiplying the expression by \( \frac{1}{2} \times 2 \) to convert the quadratic part into a sum of squared differences \( (x-y)^2 + (y-z)^2 + (z-x)^2 \).

Question 13. If \( x + y + z = 0 \), show that \( x^3 + y^3 + z^3 = 3xyz \).
Answer:
We know the identity: \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \).
Given that \( x + y + z = 0 \).
Substitute this into the identity:
\( x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx) \)
Since anything multiplied by zero is zero:
\( x^3 + y^3 + z^3 - 3xyz = 0 \)
Move \( -3xyz \) to the other side:
\( x^3 + y^3 + z^3 = 3xyz \)
Hence, it is shown.
In simple words: We start with the main rule for \( x^3+y^3+z^3-3xyz \). If \( x+y+z \) equals zero, then the whole right side of the equation becomes zero because anything multiplied by zero is zero. This leaves us with just \( x^3+y^3+z^3 = 3xyz \), which is what we needed to prove.

Exam Tip: This is a special condition of the identity \( a^3+b^3+c^3-3abc \). Always state the general identity first, then apply the given condition to simplify and prove the required result.

Question 14. Without actually calculating the cubes, find the value of each of the following:
(i) \( (-12)^3 + (7)^3 + (5)^3 \)
(ii) \( (28)^3 + (-15)^3 + (-13)^3 \)
Answer:
(i) Let \( a = -12 \), \( b = 7 \), and \( c = 5 \).
First, check the sum \( a + b + c \):
\( a + b + c = -12 + 7 + 5 \)
\( = -12 + 12 \)
\( = 0 \)
We know that if \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \).
So, \( (-12)^3 + (7)^3 + (5)^3 \)
\( = 3 \times (-12) \times 7 \times 5 \)
\( = 3 \times (-12) \times 35 \)
\( = -36 \times 35 \)
\( = -1260 \)
In simple words: We check if the three numbers add up to zero. If they do, there's a special shortcut: the sum of their cubes is just three times their product. So we multiply 3 by -12, 7, and 5 to get the answer.
(ii) Let \( a = 28 \), \( b = -15 \), and \( c = -13 \).
First, check the sum \( a + b + c \):
\( a + b + c = 28 + (-15) + (-13) \)
\( = 28 - 15 - 13 \)
\( = 28 - 28 \)
\( = 0 \)
Since \( a + b + c = 0 \), we use the identity \( a^3 + b^3 + c^3 = 3abc \).
So, \( (28)^3 + (-15)^3 + (-13)^3 \)
\( = 3 \times 28 \times (-15) \times (-13) \)
\( = 3 \times 28 \times (195) \)
\( = 84 \times 195 \)
\( = 16380 \)
In simple words: Again, we add the three given numbers together. Since they sum to zero, we can use the same special rule: the sum of their cubes is 3 multiplied by the product of the numbers. We just multiply 3, 28, -15, and -13 to get the answer.

Exam Tip: This question directly applies the special case of the identity from Question 13. Always check if the sum of the three numbers is zero before applying the formula \( a^3+b^3+c^3 = 3abc \).

Question 15. Give possible expressions for the length and breadth of each of the following rectangles in which their areas are given:
(i) Area: \( 25a^2 – 35a + 12 \)
(ii) Area: \( 35y^2 + 13y – 12 \)
Answer:
(i) The area of a rectangle is given by Length \( \times \) Breadth. We need to factorise the quadratic expression for the area.
Area: \( 25a^2 - 35a + 12 \)
We use splitting the middle term. We need two numbers that multiply to \( 25 \times 12 = 300 \) and add up to \( -35 \). These numbers are \( -20 \) and \( -15 \).
\( = 25a^2 - 20a - 15a + 12 \)
Group terms and factor out common factors:
\( = 5a(5a - 4) - 3(5a - 4) \)
\( = (5a - 4)(5a - 3) \)
Therefore, the possible expressions for the length and breadth of the rectangle are \( (5a - 3) \) and \( (5a - 4) \).
In simple words: To find the length and breadth, we need to break the area expression into two multiplication parts. We do this by splitting the middle number (-35a) into two parts that add to -35a and multiply to 25a\(^2\) times 12. Then we group and factor to get the two expressions.
(ii) Area: \( 35y^2 + 13y - 12 \)
We use splitting the middle term. We need two numbers that multiply to \( 35 \times (-12) = -420 \) and add up to \( 13 \). These numbers are \( 28 \) and \( -15 \).
\( = 35y^2 + 28y - 15y - 12 \)
Group terms and factor out common factors:
\( = 7y(5y + 4) - 3(5y + 4) \)
\( = (5y + 4)(7y - 3) \)
Therefore, the possible expressions for the length and breadth of the rectangle are \( (7y - 3) \) and \( (5y + 4) \).
In simple words: We factorise this area expression similarly. We find two numbers that multiply to 35y\(^2\) times -12 and add up to 13y. We split the middle term, group, and factor out common terms to find the length and breadth.

Exam Tip: For factorising quadratic expressions like \( ax^2 + bx + c \), always use the "splitting the middle term" method. Find two numbers that multiply to \( ac \) and add up to \( b \).

Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: \( 3x^2 – 12x \)
(ii) Volume: \( 12ky^2 + 8ky – 20k \)
Answer:
(i) The volume of a cuboid is Length \( \times \) Breadth \( \times \) Height. We need to factorise the given expression into three factors.
Volume: \( 3x^2 - 12x \)
First, factor out the common term, which is \( 3x \).
\( = 3x(x - 4) \)
Therefore, the possible expressions for the dimensions of the cuboid are \( 3 \), \( x \), and \( (x - 4) \).
In simple words: To find the three dimensions, we need to break the volume expression into three things multiplied together. We find the greatest common factor, which is 3x. When we factor that out, we are left with \( (x-4) \), giving us the three dimensions.
(ii) Volume: \( 12ky^2 + 8ky - 20k \)
First, factor out the common term, which is \( 4k \).
\( = 4k(3y^2 + 2y - 5) \)
Now, we need to factorise the quadratic expression inside the parenthesis \( (3y^2 + 2y - 5) \).
We use splitting the middle term. We need two numbers that multiply to \( 3 \times (-5) = -15 \) and add up to \( 2 \). These numbers are \( 5 \) and \( -3 \).
\( = 4k(3y^2 + 5y - 3y - 5) \)
Group terms and factor out common factors within the parenthesis:
\( = 4k[y(3y + 5) - 1(3y + 5)] \)
\( = 4k(3y + 5)(y - 1) \)
Therefore, the possible expressions for the dimensions of the cuboid are \( 4k \), \( (3y + 5) \), and \( (y - 1) \).
In simple words: We first pull out the common factor, which is 4k. Then, we factor the remaining part, which is a quadratic expression. We split its middle term (2y) into two parts that multiply to \( 3y^2 \) times -5 and add to 2y. This gives us two more factors, making a total of three dimensions.

Exam Tip: For volume factorisation, always look for a common factor first. Then, if a quadratic expression remains, factorise it using the "splitting the middle term" method.