Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.4

Question 1. Find the following products?

i. (3x + 2y) (9x² – 6xy + 4y²)

Solution:

We know that  a³ + b³ = (a + b)(a² – ab + b²) 

we can write the given equation as,

=> (3x + 2y)[(3x)² – 6xy + (2y)²]

=> (3x)³ + (2y)³

=> 27x³ +  8y³

ii. (4x – 5y) (16x² + 20xy  +  25y²)

Solution:

We know that a³ – b³ = (a – b)(a² + ab + b²)

we can write the given equation as,

=> (4x – 5y)[(4x)² + 20xy + (5y)²]

=> (4x)³ – (5y)³

=> 64x³ – 125y³

iii. (7p⁴ + q) (49p⁸ – 7p⁴q + q²)

Solution:

We can write the given equation as,

=> (7p⁴ + q)[(7p⁴)² – 7p⁴q + q²]

We know that  a³  + b³= (a + b)(a² – ab + b²) 

=> (7p⁴)³ + q³

=> 343p¹² + q³

iv. [(x/2) + 2y] [(x²/4) – xy + 4y²]

Solution:

We can write the given equation as,

[(x / 2) + 2y] [(x / 2)²  – (x / 2) * 2y + (2y)²]  —— eq(i)

By writing the given equation as eq(i)  we can easily make the equation as  (a + b)[a² – ab + b²] = a³ + b³

So, the above e quation can be solved as,

=> (x / 2)³ + (2y)³

=> (x³ / 8) + 8y³

v. [(3/x) – (5/y)] [(9/x²) + (25/y²) + (15/xy)]

Solution:

We can write given equation as,

[(3 / x) – (5 / y)] {(3 / x)² + (3 / x)(5 / y) + (5 / y)²]

So, above equation makes the identity of a³ – b³

Now,

=> (3 / x)³ – (5 / y)³

=> (27 / x³) – (125 / y³)

vi. [3 + (5/x)] [9 – (15/x) + ( 25/x²)]

Solution:

We can write the given equation as,

=> [3 + (5 / x)] [(3)² – 3 * (5 / x) + (5 / x)²]

So, above equation makes the identity of a³ + b³

=> (3)³ + (5 / x)³

=> 27 + (125 / x²)

vii. [(2/x) + 3x] [(4/x²) + 9x² – 6)]

Solution:

We can write the given equation as,

=> [(2 / x) + 3x]  [(2 / x)²  – (2 / x)(3x) + (3x)²]

So, above equation makes the identity of a³ + b³

=> (2 / x)³ + (3x)³

=> (8 / x³) + 27x³

viii. [(3/2) – 2x²] [(9/x²) + 4x⁴  –  6x]

Solution:

We can write the given equation as, 

=> [(3 / x) – 2x²] [(3 / x)² – (3 / x)(2x²) + (2x²)²]

So, above equation makes the identity of a³ – b³

=> (3 / x)³ – (2x²)³

=> (27 / x³) –  8x⁶

ix. (1 – x)(1 + x + x²)

Solution:

This equation is clearly making the identity of a³ – b³

=> 1³ – x³

=> 1 – x³

x. (1 + x)(1 – x + x²)

Solution:

This equation is clearly making the identity of a³ + b³

=> 1³ + x³

=> 1 + x³

xi. (x² – 1)(x⁴ + x² + 1)

Solution:

We can write the given equation as,

=> (x²  – 1 ) [(x²)²  + x²   + 1)]

This equation is clearly making the identity of a³  –  b³

=> (x²)³  –  1 

=> x⁶ – 1

xii. (x³ + 1)(x⁶ – x³  + 1)

Solution:

We can write the given equation as,

=> (x³ + 1) [(x³)² – x³ + 1]

This equation is clearly making the identity of a³ + b³

=> (x³)³ + 1

=> x⁹ + 1

Question 2. If x = 3 and y = -1, find the values of each of the following using in identity?

i. (9y² – 4x²)  (81y⁴ + 36x²y²  + 16x⁴)

Solution:

We can write the given equation as,

=> (9y²  – 4x²) [(9y²)² +  9y² * 4x² + (4x²)²]

This is now clearly making the identity of a³ – b³

=> (9y²)³  –  (4x²)³

=> 729y⁶  –  64x⁶   —–eq(i)

Putting the given values in eq(i)

=> 729 * 1  –  64 * 729

=> 729  –  46656

=> -45927

ii. [(3/x) – (x/3)] [(x²/9) + (9/x²) + 1]

Solution:

We can write the given equation as,

=> [(3 / x) – (x / 3)]  [(x / 3)² + (x / 3)(3 / x) + (3 / x)²]

This is making the identity of a³ – b³

=> (3 / x)³ – (x / 3)³   —-eq(i)

Putting the given values in eq(i)

=> 1 – 1

=> 0

iii. [(x/7) + (y/3)] [( x²/49) + (y²/9) – (xy/21)]

Solution:

We can write the given equation as,

=> [(x / 7) + (y / 3)]  [(x / 7)²  – (x / 7)(y / 3) – (y / 3 )²]

This is making the identity of a³ + b³

=> (x / 7)³ + (y / 3)³    —eq(i)

Putting the values in eq(i)

=> 27 / 343 – 1 / 27

=> (729 – 343) / 9261

=> 386 / 9261

iv. [(x/4) – (y/3)] [(x²/16) + (xy/12) + (y²/9)]

Solution:

We can write this equation as,

=> [(x / 4) – (y / 3)] [(x / 4)² + (x / 4)(y / 3) + (y / 3)²]

This is clearly making the identity of a³ – b³

=> (x / 4)³  – (y / 3)³

=> (x³ / 64) – (y³ / 27) —eq(i)

Putting the values in eq(i)

=> (27 / 64) + (1 / 27)

=> (729 + 64) / 1728

=> 793 / 1728

Question 3. If a + b = 10 and ab = 16, find the value of a² – ab + b² and a² + ab + b²?

Solution:

Taking a + b = 10 

On squaring both sides,

=> (a + b)² = (10)²

We get, a² + b² + 2ab = 100   —eq(i)

Putting the value of ab = 16 in eq(i)

=> a² + b² +  2 * 16 = 100

=> a² + b² + 32 =100

=> a² + b² = 100 – 32 = 68

So, a² – ab + b² = a² + b² – ab = 68 – 16 = 52

and a² + ab + b² = a² + b² + ab = 68 + 16 = 84

Question 4. If a + b = 8 and ab = 6, find the value of a³ + b³?

Solution:

Taking a + b = 8

On cubing both sides,

(a + b)³ = (8)³

=> a³ + b³ + 3ab(a + b) = 512 —-eq(i)

Putting the given values in eq(i)

=> a³ + b³ + 3 * 6 * 8 = 512

=> a³ + b³ + 144 = 512

=> a³ + b³ = 512 – 144 = 368

=> a³ + b³  = 368

Question 5. If a – b = 6 and ab = 20 , find the value of a³ – b³?

Solution:

Taking a – b=6

On cubing both sides,

(a – b)³ = (6)³

=> a³ – b³ – 3ab(a – b) = 216   —eq(i)

Putting the given values in eq(i)

=> a³ – b³ – 3 * 20 * 6 = 216

=> a³ – b³ – 360 = 216

=> a³ – b³ = 216 + 360 = 576

=> a³ – b³ = 576

Question 6. If x = -2 and y = 1, by using an identity find the value of the following:

i. (4y² – 9x²)(16y⁴ + 36x²y² + 81x⁴)

Solution:

Given equation can be written as,

=> (4y² – 9x²)[(4y²)² + 4y² * 9x² + (9x²)²]

This equation now making the identity of a³ – b³

=> (4y²)³ – (9x²)³

=> 64y⁶  – 729x⁶      —–eq(i)

Putting the given values  in eq(i)

=> 64 * 1⁶  – 729 * (-2)⁶

=> 64 – 729 * 64

=> 64 – 46656

=> -46592

ii. [(2/x) – (x/2)][(4/x²) + (x²/4) + 1]

Solution:

We can write this equation as,

=> [(2 / x) – (x / 2)] [(2 / x)² + 2(2 / x)(x / 2) + (x / 2)²]

This equation is clearly making the identity of a³ – b³

=> (2 / x)³ – (x / 2)³

=> (8 / x³) – (x³ / 8) —eq(i)

Putting the given values in eq(i)

=> [8 / (-2)³] – [(-2)³ / 8]

=> -1 + 1

=> 0