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

### Question 1. Find the following products?

**i. (3x + 2y) (9x ^{2} – 6xy + 4y^{2})**

**Solution:**

We know that a

^{3 }+ b^{3 }= (a + b)(a^{2 }– ab + b^{2})we can write the given equation as,

=> (3x + 2y)[(3x)

^{2 }– 6xy + (2y)^{2}]=> (3x)

^{3 }+ (2y)^{3}=>

27x^{3 }+ 8y^{3}

**ii. (4x – 5y) (16x ^{2 }+ 20xy + 25y^{2})**

**Solution:**

We know that a

^{3 }– b^{3 }= (a – b)(a^{2 }+ ab + b^{2})we can write the given equation as,

=> (4x – 5y)[(4x)

^{2 }+ 20xy + (5y)^{2}]=> (4x)

^{3 }– (5y)^{3}=>

64x^{3 }– 125y^{3}

**iii. (7p ^{4 }+ q) (49p^{8 }– 7p^{4}q + q^{2})**

**Solution:**

We can write the given equation as,

=> (7p

^{4 }+ q)[(7p^{4})^{2 }– 7p^{4}q + q^{2}]We know that a

^{3 }+ b^{3}= (a + b)(a^{2 }– ab + b^{2})=> (7p

^{4})^{3 }+ q^{3}=>

343p^{12 }+ q^{3}

**iv. [(x/2) + 2y] [(x ^{2}/4) – xy + 4y^{2}]**

**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)[a

^{2 }– ab + b^{2}] = a^{3 }+ b^{3}So, the above e quation can be solved as,

=> (x / 2)

^{3 }+ (2y)^{3}=>

(x^{3 }/ 8) + 8y^{3}

**v. [(3/x) – (5/y)] [(9/x ^{2}) + (25/y^{2}) + (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 a

^{3 }– b^{3}Now,

=> (3 / x)

^{3 }– (5 / y)^{3}=>

(27 / x^{3}) – (125 / y^{3})

**vi. [3 + (5/x)] [9 – (15/x) + ( 25/x ^{2})]**

**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 a

^{3 }+ b^{3}=> (3)

^{3 }+ (5 / x)^{3}=>

27 + (125 / x^{2})

**vii. [(2/x) + 3x] [(4/x ^{2}) + 9x^{2 }– 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 a

^{3 }+ b^{3}=> (2 / x)

^{3 }+ (3x)^{3}=>

(8 / x^{3}) + 27x^{3}

**viii. [(3/2) – 2x ^{2}] [(9/x^{2}) + 4x^{4 }– 6x]**

**Solution:**

We can write the given equation as,

=> [(3 / x) – 2x

^{2}] [(3 / x)^{2 }– (3 / x)(2x^{2}) + (2x^{2})^{2}]So, above equation makes the identity of a

^{3 }– b^{3}=> (3 / x)

^{3 }– (2x^{2})^{3}=>

(27 / x^{3})^{ }–^{ }8x^{6}

**ix. (1 – x)(1 + x + x ^{2})**

**Solution:**

This equation is clearly making the identity of a

^{3 }– b^{3}=> 1

^{3 }– x^{3}=>

1 – x^{3}

**x. (1 + x)(1 – x + x ^{2})**

**Solution:**

This equation is clearly making the identity of a

^{3 }+ b^{3}=> 1

^{3 }+ x^{3}=>

1 + x^{3}

**xi. (x ^{2 }– 1)(x^{4 }+ x^{2 }+ 1)**

**Solution:**

We can write the given equation as,

=> (x

^{2 }– 1 ) [(x^{2})^{2 }+ x^{2 }+ 1)]This equation is clearly making the identity of a

^{3 }– b^{3}=> (x

^{2})^{3 }– 1=>

x^{6 }– 1

**xii. (x ^{3 }+ 1)(x^{6 }– x^{3 }+ 1)**

**Solution:**

We can write the given equation as,

=> (x

^{3 }+ 1) [(x^{3})^{2 }– x^{3 }+ 1]This equation is clearly making the identity of a

^{3 }+ b^{3}=> (x

^{3})^{3 }+ 1=>

x^{9 }+ 1

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

**i. (9y ^{2 }– 4x^{2}) (81y^{4 }+ 36x^{2}y^{2 }+ 16x^{4})**

**Solution:**

We can write the given equation as,

=> (9y

^{2 }– 4x^{2}) [(9y^{2})^{2 }+ 9y^{2 }* 4x^{2 }+ (4x^{2})^{2}]This is now clearly making the identity of a

^{3 }– b^{3}=> (9y

^{2})^{3 }– (4x^{2})^{3}=> 729y

^{6 }– 64x^{6 }—–eq(i)Putting the given values in eq(i)

=> 729 * 1 – 64 * 729

=> 729 – 46656

=>

-45927

**ii. [(3/x) – (x/3)] [(x ^{2}/9) + (9/x^{2}) + 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 a

^{3 }– b^{3}=> (3 / x)

^{3 }– (x / 3)^{3}—-eq(i)Putting the given values in eq(i)

=> 1 – 1

=>

0

**iii. [(x/7) + (y/3)] [( x ^{2}/49) + (y^{2}/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 a

^{3 }+ b^{3}=> (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)] [(x ^{2}/16) + (xy/12) + (y^{2}/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 a

^{3 }– b^{3}=> (x / 4)

^{3 }– (y / 3)^{3}=> (x

^{3 }/ 64) – (y^{3 }/ 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 a^{2 }– ab + b^{2} and a^{2 }+ ab + b^{2}?

**Solution:**

Taking a + b = 10

On squaring both sides,

=> (a + b)

^{2 }= (10)^{2}We get, a

^{2 }+ b^{2 }+ 2ab = 100 —eq(i)Putting the value of ab = 16 in eq(i)

=> a

^{2 }+ b^{2 }+ 2 * 16 = 100=> a

^{2 }+ b^{2 }+^{ }32 =100=> a

^{2 }+ b^{2 }= 100 – 32 = 68

So, a^{2 }– ab + b^{2 }= a^{2 }+ b^{2 }– ab = 68 – 16 = 52

and a^{2 }+ ab + b^{2 }= a^{2 }+ b^{2 }+ ab = 68 + 16 = 84

### Question 4. If a + b = 8 and ab = 6, find the value of a^{3 }+ b^{3}?

**Solution:**

Taking a + b = 8

On cubing both sides,

(a + b)

^{3 }= (8)^{3}=> a

^{3 }+ b^{3 }+ 3ab(a + b) = 512 —-eq(i)Putting the given values in eq(i)

=> a

^{3 }+ b^{3 }+ 3 * 6 * 8 = 512=> a

^{3 }+ b^{3 }+ 144 = 512=> a

^{3 }+ b^{3 }= 512 – 144 = 368=>

a^{3 }+ b^{3 }= 368

### Question 5. If a – b = 6 and ab = 20 , find the value of a^{3 }– b^{3}?

**Solution:**

Taking a – b=6

On cubing both sides,

(a – b)

^{3 }= (6)^{3}=> a

^{3 }– b^{3 }– 3ab(a – b) = 216 —eq(i)Putting the given values in eq(i)

=> a

^{3 }– b^{3 }– 3 * 20 * 6 = 216=> a

^{3 }– b^{3 }– 360 = 216=> a

^{3 }– b^{3 }= 216 + 360 = 576=>

a^{3 }– b^{3 }= 576

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

**i. (4y ^{2 }– 9x^{2})(16y^{4 }+ 36x^{2}y^{2 }+ 81x^{4})**

**Solution:**

Given equation can be written as,

=> (4y

^{2 }– 9x^{2})[(4y^{2})^{2 }+ 4y^{2 }* 9x^{2 }+ (9x^{2})^{2}]This equation now making the identity of a

^{3 }– b^{3}=> (4y

^{2})^{3 }– (9x^{2})^{3}=> 64y

^{6 }– 729x^{6}—–eq(i)Putting the given values in eq(i)

=> 64 * 1

^{6 }– 729 * (-2)^{6}=> 64 – 729 * 64

=> 64 – 46656

=>

-46592

**ii. [(2/x) – (x/2)][(4/x ^{2}) + (x^{2}/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 a

^{3 }– b^{3}=> (2 / x)

^{3 }– (x / 2)^{3}=> (8 / x

^{3}) – (x^{3 }/ 8) —eq(i)Putting the given values in eq(i)

=> [8 / (-2)

^{3}] – [(-2)^{3 }/ 8]=> -1 + 1

=>

0