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

  • Difficulty Level : Expert
  • Last Updated : 15 Dec, 2020
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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  –  64x  —–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

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