OP Malhotra Class 9 Maths Solutions Chapter 3 Expansions Exercise 3 (B)

Question 1. Find the value of:
(i) \( a^2 + b^2 \) when \( a + b = 9, ab = 20 \)
(ii) \( p^2 + q^2 \) if \( p - q = 6 \) and \( p + q = 14 \)
(iii) \( mn \) if \( m + n = 8, m - n = 2 \)
(iv) \( x^2 + \frac{1}{x^2} \) and \( x^4 + \frac{1}{x^4} \) if \( x + \frac{1}{x} = 3 \)
(v) \( x^2 - \frac{1}{x^2} \) if \( x + \frac{1}{x} = \sqrt{5} \)
Answer:
(i) We are given \( a + b = 9 \) and \( ab = 20 \). We need to find the value of \( a^2 + b^2 \).
We use the identity: \( a^2 + b^2 = (a+b)^2 - 2ab \)
Now, substitute the given values:
\( a^2 + b^2 = (9)^2 - 2 \times 20 \)
\( = 81 - 40 \)
\( = 41 \)
So, the value of \( a^2 + b^2 \) is \( 41 \). This identity helps us find the sum of squares when the sum and product of numbers are known.
(ii) We are given \( p - q = 6 \) and \( p + q = 14 \). We need to find the value of \( p^2 + q^2 \).
We use the identity: \( 2(p^2 + q^2) = (p+q)^2 + (p-q)^2 \)
Now, substitute the given values:
\( 2(p^2 + q^2) = (14)^2 + (6)^2 \)
\( = 196 + 36 \)
\( = 232 \)
To find \( p^2 + q^2 \), divide by 2:
\( p^2 + q^2 = \frac{232}{2} \)
\( = 116 \)
Thus, the value of \( p^2 + q^2 \) is \( 116 \). This shows how the sum of squares relates to the sum and difference of the numbers.
(iii) We are given \( m + n = 8 \) and \( m - n = 2 \). We need to find the value of \( mn \).
We use the identity: \( 4mn = (m+n)^2 - (m-n)^2 \)
Now, substitute the given values:
\( 4mn = (8)^2 - (2)^2 \)
\( = 64 - 4 \)
\( = 60 \)
To find \( mn \), divide by 4:
\( mn = \frac{60}{4} \)
\( = 15 \)
Therefore, the value of \( mn \) is \( 15 \). This identity is very useful for finding the product of two numbers when their sum and difference are known.
(iv) We are given \( x + \frac{1}{x} = 3 \). We need to find \( x^2 + \frac{1}{x^2} \) and \( x^4 + \frac{1}{x^4} \).
First, for \( x^2 + \frac{1}{x^2} \):
Square both sides of the given equation:
\( \left(x + \frac{1}{x}\right)^2 = (3)^2 \)
Use the identity \( (a+b)^2 = a^2+2ab+b^2 \):
\( x^2 + 2 \cdot x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2 = 9 \)
\( x^2 + 2 + \frac{1}{x^2} = 9 \)
Move the 2 to the other side:
\( \implies x^2 + \frac{1}{x^2} = 9 - 2 \)
\( = 7 \)
Next, for \( x^4 + \frac{1}{x^4} \):
Square both sides of the result \( x^2 + \frac{1}{x^2} = 7 \):
\( \left(x^2 + \frac{1}{x^2}\right)^2 = (7)^2 \)
Use the identity \( (a+b)^2 = a^2+2ab+b^2 \):
\( (x^2)^2 + 2 \cdot x^2 \cdot \frac{1}{x^2} + \left(\frac{1}{x^2}\right)^2 = 49 \)
\( x^4 + 2 + \frac{1}{x^4} = 49 \)
Move the 2 to the other side:
\( \implies x^4 + \frac{1}{x^4} = 49 - 2 \)
\( = 47 \)
So, \( x^2 + \frac{1}{x^2} = 7 \) and \( x^4 + \frac{1}{x^4} = 47 \). Repeated squaring helps to find higher powers of expressions like these.
(v) We are given \( x + \frac{1}{x} = \sqrt{5} \). We need to find \( x^2 - \frac{1}{x^2} \).
First, find \( x^2 + \frac{1}{x^2} \):
Square both sides of the given equation:
\( \left(x + \frac{1}{x}\right)^2 = (\sqrt{5})^2 \)
\( x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 5 \)
\( x^2 + 2 + \frac{1}{x^2} = 5 \)
\( \implies x^2 + \frac{1}{x^2} = 5 - 2 \)
\( = 3 \)
Next, find \( x - \frac{1}{x} \):
We know that \( \left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \)
Substitute the value of \( x^2 + \frac{1}{x^2} \):
\( \left(x - \frac{1}{x}\right)^2 = 3 - 2 \)
\( = 1 \)
Take the square root of both sides:
\( \implies x - \frac{1}{x} = \pm 1 \)
Finally, find \( x^2 - \frac{1}{x^2} \):
We use the difference of squares identity: \( a^2 - b^2 = (a+b)(a-b) \)
Here, \( a=x \) and \( b=\frac{1}{x} \).
\( x^2 - \frac{1}{x^2} = \left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right) \)
Substitute the known values:
\( x^2 - \frac{1}{x^2} = \sqrt{5} \times (\pm 1) \)
\( \implies x^2 - \frac{1}{x^2} = \pm \sqrt{5} \)
So, \( x^2 - \frac{1}{x^2} \) can be \( \sqrt{5} \) or \( -\sqrt{5} \). This problem combines several algebraic identities to reach the solution.
In simple words: For (i), (ii), (iii), use common algebraic formulas that connect sums, products, and differences to squares. For (iv) and (v), take the given equation and square both sides to find the first part, then square again for the second part if needed, or use other identities like difference of squares. Remember that squaring can sometimes give positive and negative results.

🎯 Exam Tip: Always recall the basic algebraic identities like \( (a+b)^2 \), \( (a-b)^2 \), \( (a+b+c)^2 \), \( a^2-b^2 \), \( (a+b)^3 \), and \( (a-b)^3 \) as they are fundamental for solving expansion problems. Pay close attention to signs when taking square roots.

Question 2.
(i) \( a^2 + b^2 + c^2 \) if \( a + b + c = 17 \) and \( ab + bc + ca = 30 \).
(ii) \( ab + bc + ca \) if \( a + b + c = 15 \) and \( a^2 + b^2 + c^2 = 77 \).
(iii) \( a + b + c \) if \( a^2 + b^2 + c^2 = 50 \) and \( ab + bc + ca = 47 \).
Answer:
(i) We are given \( a + b + c = 17 \) and \( ab + bc + ca = 30 \). We need to find \( a^2 + b^2 + c^2 \).
We use the identity: \( (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) \)
Substitute the given values into the identity:
\( (17)^2 = a^2+b^2+c^2 + 2(30) \)
\( 289 = a^2+b^2+c^2 + 60 \)
To find \( a^2+b^2+c^2 \), subtract 60 from 289:
\( \implies a^2+b^2+c^2 = 289 - 60 \)
\( = 229 \)
So, the value of \( a^2+b^2+c^2 \) is \( 229 \). This identity is key for problems involving three variables.
(ii) We are given \( a + b + c = 15 \) and \( a^2 + b^2 + c^2 = 77 \). We need to find \( ab + bc + ca \).
We use the identity: \( (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) \)
Substitute the given values:
\( (15)^2 = 77 + 2(ab+bc+ca) \)
\( 225 = 77 + 2(ab+bc+ca) \)
Subtract 77 from 225:
\( \implies 2(ab+bc+ca) = 225 - 77 \)
\( = 148 \)
To find \( ab+bc+ca \), divide by 2:
\( \implies ab+bc+ca = \frac{148}{2} \)
\( = 74 \)
Therefore, the value of \( ab+bc+ca \) is \( 74 \). This shows how to find the sum of products from the sum and sum of squares.
(iii) We are given \( a^2 + b^2 + c^2 = 50 \) and \( ab + bc + ca = 47 \). We need to find \( a + b + c \).
We use the identity: \( (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) \)
Substitute the given values:
\( (a+b+c)^2 = 50 + 2(47) \)
\( = 50 + 94 \)
\( = 144 \)
To find \( a+b+c \), take the square root of 144:
\( \implies a+b+c = \pm \sqrt{144} \)
\( = \pm 12 \)
So, \( a+b+c \) can be \( 12 \) or \( -12 \). Remember that the square root of a number can be both positive and negative.
In simple words: For all parts, use the main formula that links the square of the sum of three numbers to their individual squares and their pair products. Rearrange the formula to find the missing part by plugging in the numbers you already know. Always remember that taking a square root gives two possible answers, one positive and one negative.

🎯 Exam Tip: When finding a variable by taking a square root (like \( a+b+c \) in part iii), always remember to include both the positive and negative solutions to get full marks.

Question 3.
(i) \( 8x^3 + 84x^2y + 294xy^2 + 343y^3 \) if \( x = 1, y = 2 \)
(ii) \( 27x^3 - 27x^2y + 9xy^2 - y^3 \) if \( x = 2, y = 1 \)
Answer:
(i) We need to evaluate the expression \( 8x^3 + 84x^2y + 294xy^2 + 343y^3 \) when \( x = 1 \) and \( y = 2 \).
This expression looks like the expansion of a cube sum, \( (A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3 \).
Let's rewrite the terms to match this pattern:
\( 8x^3 = (2x)^3 \)
\( 343y^3 = (7y)^3 \)
Now, check the middle terms:
\( 3(2x)^2(7y) = 3(4x^2)(7y) = 84x^2y \) (Matches)
\( 3(2x)(7y)^2 = 3(2x)(49y^2) = 294xy^2 \) (Matches)
So, the expression is \( (2x+7y)^3 \).
Now substitute the values \( x=1 \) and \( y=2 \):
\( (2 \times 1 + 7 \times 2)^3 \)
\( = (2 + 14)^3 \)
\( = (16)^3 \)
\( = 4096 \)
Thus, the value of the expression is \( 4096 \). Recognizing the cubic identity simplifies the calculation significantly.
(ii) We need to evaluate the expression \( 27x^3 - 27x^2y + 9xy^2 - y^3 \) when \( x = 2 \) and \( y = 1 \).
This expression looks like the expansion of a cube difference, \( (A-B)^3 = A^3 - 3A^2B + 3AB^2 - B^3 \).
Let's rewrite the terms to match this pattern:
\( 27x^3 = (3x)^3 \)
\( y^3 = (y)^3 \)
Now, check the middle terms:
\( -3(3x)^2(y) = -3(9x^2)(y) = -27x^2y \) (Matches)
\( 3(3x)(y)^2 = 3(3x)(y^2) = 9xy^2 \) (Matches)
So, the expression is \( (3x-y)^3 \).
Now substitute the values \( x=2 \) and \( y=1 \):
\( (3 \times 2 - 1)^3 \)
\( = (6 - 1)^3 \)
\( = (5)^3 \)
\( = 125 \)
Hence, the value of the expression is \( 125 \). Applying the correct algebraic identity is crucial here.
In simple words: For both parts, first look at the expression and see if it looks like the expanded form of \( (A+B)^3 \) or \( (A-B)^3 \). Figure out what A and B are. Then, replace the whole expression with its simpler cubic form like \( (A+B)^3 \). Finally, put the given values for x and y into this simpler form and calculate the answer.

🎯 Exam Tip: Practice recognizing the patterns for \( (a+b)^3 \) and \( (a-b)^3 \) expansions. This allows you to condense the expression into a simpler cubic form before substituting values, which reduces calculation errors.

Question 4.
(i) \( a^3 + b^3 \) if \( a + b = 3 \) and \( ab = 2 \)
(ii) \( x^3 + \frac{1}{x^3} \) if \( (x + \frac{1}{x}) = 3 \)
(iii) \( x^3 - \frac{1}{x^3} \) if \( x^2 + \frac{1}{x^2} = 18 \)
(iv) \( x^3 + \frac{1}{125x^3} \) if \( x^2 + \frac{1}{25x^2} = 8 \frac{3}{5} \)
Answer:
(i) We are given \( a + b = 3 \) and \( ab = 2 \). We need to find \( a^3 + b^3 \).
We use the identity: \( a^3 + b^3 = (a+b)^3 - 3ab(a+b) \)
Now, substitute the given values:
\( a^3 + b^3 = (3)^3 - 3(2)(3) \)
\( = 27 - 18 \)
\( = 9 \)
Thus, the value of \( a^3 + b^3 \) is \( 9 \). This identity is a shortcut to finding the sum of cubes.
(ii) We are given \( x + \frac{1}{x} = 3 \). We need to find \( x^3 + \frac{1}{x^3} \).
We use the identity: \( x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \)
Now, substitute the given value:
\( x^3 + \frac{1}{x^3} = (3)^3 - 3(3) \)
\( = 27 - 9 \)
\( = 18 \)
So, the value of \( x^3 + \frac{1}{x^3} \) is \( 18 \). This formula helps solve specific reciprocal cubic expressions easily.
(iii) We are given \( x^2 + \frac{1}{x^2} = 18 \). We need to find \( x^3 - \frac{1}{x^3} \).
First, we need to find \( x - \frac{1}{x} \).
We know the identity: \( \left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \)
Substitute the given value for \( x^2 + \frac{1}{x^2} \):
\( \left(x - \frac{1}{x}\right)^2 = 18 - 2 \)
\( = 16 \)
Take the square root of both sides:
\( \implies x - \frac{1}{x} = \pm \sqrt{16} \)
\( \implies x - \frac{1}{x} = \pm 4 \)
Now, we use the identity for \( x^3 - \frac{1}{x^3} \): \( x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right) \)
We have two possible cases for \( x - \frac{1}{x} \):
Case 1: If \( x - \frac{1}{x} = 4 \)
\( x^3 - \frac{1}{x^3} = (4)^3 + 3(4) \)
\( = 64 + 12 \)
\( = 76 \)
Case 2: If \( x - \frac{1}{x} = -4 \)
\( x^3 - \frac{1}{x^3} = (-4)^3 + 3(-4) \)
\( = -64 - 12 \)
\( = -76 \)
Therefore, the value of \( x^3 - \frac{1}{x^3} \) is \( \pm 76 \). We must consider both positive and negative solutions for \( x - \frac{1}{x} \).
(iv) We are given \( x^2 + \frac{1}{25x^2} = 8 \frac{3}{5} \). We need to find \( x^3 + \frac{1}{125x^3} \).
First, convert the mixed fraction: \( 8 \frac{3}{5} = \frac{8 \times 5 + 3}{5} = \frac{40+3}{5} = \frac{43}{5} \)
So, the given equation is \( x^2 + \frac{1}{25x^2} = \frac{43}{5} \). This can be written as \( x^2 + \left(\frac{1}{5x}\right)^2 = \frac{43}{5} \).
To find \( x + \frac{1}{5x} \), we can complete the square for \( \left(x + \frac{1}{5x}\right)^2 \).
We know \( \left(x + \frac{1}{5x}\right)^2 = x^2 + \left(\frac{1}{5x}\right)^2 + 2 \cdot x \cdot \frac{1}{5x} \)
\( = x^2 + \frac{1}{25x^2} + \frac{2}{5} \)
Substitute the given value:
\( \left(x + \frac{1}{5x}\right)^2 = \frac{43}{5} + \frac{2}{5} \)
\( = \frac{45}{5} = 9 \)
Take the square root of both sides:
\( \implies x + \frac{1}{5x} = \pm \sqrt{9} \)
\( \implies x + \frac{1}{5x} = \pm 3 \)
Now, we need to find \( x^3 + \frac{1}{125x^3} \). This can be written as \( x^3 + \left(\frac{1}{5x}\right)^3 \).
We use the identity for \( A^3+B^3 = (A+B)^3 - 3AB(A+B) \). Here, \( A=x \) and \( B=\frac{1}{5x} \), so \( AB = x \cdot \frac{1}{5x} = \frac{1}{5} \).
We have two possible cases for \( x + \frac{1}{5x} \):
Case 1: If \( x + \frac{1}{5x} = 3 \)
\( x^3 + \frac{1}{125x^3} = (3)^3 - 3\left(\frac{1}{5}\right)(3) \)
\( = 27 - \frac{9}{5} \)
\( = \frac{27 \times 5 - 9}{5} \)
\( = \frac{135 - 9}{5} \)
\( = \frac{126}{5} \)
Case 2: If \( x + \frac{1}{5x} = -3 \)
\( x^3 + \frac{1}{125x^3} = (-3)^3 - 3\left(\frac{1}{5}\right)(-3) \)
\( = -27 + \frac{9}{5} \)
\( = \frac{-27 \times 5 + 9}{5} \)
\( = \frac{-135 + 9}{5} \)
\( = \frac{-126}{5} \)
Therefore, the value of \( x^3 + \frac{1}{125x^3} \) is \( \pm \frac{126}{5} \). This problem requires careful use of squaring and cubing identities.
In simple words: For (i) and (ii), use the direct formulas for \( a^3+b^3 \) or \( x^3+1/x^3 \). For (iii) and (iv), you first need to find a simpler expression like \( x-1/x \) or \( x+1/(5x) \) by squaring the given information. Remember to account for both positive and negative square roots in intermediate steps, as this can lead to multiple final answers.

🎯 Exam Tip: For problems like (iii) and (iv) where you take square roots during intermediate steps, always consider both positive and negative values. This will lead to two possible final answers for expressions like \( x^3 - \frac{1}{x^3} \) or \( x^3 + \frac{1}{125x^3} \). Be careful with mixed fractions, convert them to improper fractions first.

Question 5. Evaluate:
(i) \( 102 \times 98 \)
(ii) \( 1003^2 - 997^2 \)
(iii) \( (10)^3 - (5)^3 - (5)^3 \)
Answer:
(i) We need to evaluate \( 102 \times 98 \).
We can write this as \( (100+2)(100-2) \).
This matches the algebraic identity \( (a+b)(a-b) = a^2 - b^2 \).
Here, \( a=100 \) and \( b=2 \).
So, \( (100)^2 - (2)^2 \)
\( = 10000 - 4 \)
\( = 9996 \)
This method makes multiplying numbers close to a round figure much easier.
(ii) We need to evaluate \( 1003^2 - 997^2 \).
This matches the algebraic identity \( a^2 - b^2 = (a+b)(a-b) \).
Here, \( a=1003 \) and \( b=997 \).
So, \( (1003 + 997)(1003 - 997) \)
First, calculate the sum and difference:
\( 1003 + 997 = 2000 \)
\( 1003 - 997 = 6 \)
Now, multiply these results:
\( = 2000 \times 6 \)
\( = 12000 \)
Using the identity simplifies this large squaring and subtraction problem into simple addition, subtraction, and multiplication.
(iii) We need to evaluate \( (10)^3 - (5)^3 - (5)^3 \).
We can rewrite this expression as \( (10)^3 + (-5)^3 + (-5)^3 \).
Let \( a = 10 \), \( b = -5 \), and \( c = -5 \).
First, check the sum \( a+b+c \):
\( a+b+c = 10 + (-5) + (-5) = 10 - 5 - 5 = 0 \)
There is an identity that states if \( a+b+c=0 \), then \( a^3+b^3+c^3 = 3abc \).
Using this identity:
\( (10)^3 + (-5)^3 + (-5)^3 = 3 \times 10 \times (-5) \times (-5) \)
\( = 3 \times 10 \times 25 \)
\( = 30 \times 25 \)
\( = 750 \)
This identity provides a very quick way to solve such cubic sums when the sum of the bases is zero.
In simple words: For (i) and (ii), use the difference of squares formula, \( (a^2-b^2) = (a+b)(a-b) \). This helps you avoid big calculations by turning squares into simple additions and subtractions. For (iii), if three numbers add up to zero, then the sum of their cubes is equal to three times their product.

🎯 Exam Tip: Always look for opportunities to apply algebraic identities. They simplify complex calculations greatly. Specifically, for sums of cubes where the terms add up to zero, remember the special identity \( a^3+b^3+c^3=3abc \).