# Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.3 | Set 1

### Question 1. Find the cube of each of the following binomial expressions.

(a) (1/x+y/3)

(b) (3/x-2/x2)

(c) (2x+3/x)

(d) (4-1/3x)

Solution:

(a) (1/x+y/3)

Given, (1/x+y/3)3                                                               [(a+b)3=a3+b3+3ab(a+b)]

we know that here, a = 1/x, b= y/3

By using formula:

(1/x+y/3)3 = (1/x)3 + (y/3)3 + 3(1/x)(y/3)(1/x+y/3)

= 1/x3 +y3/27 + y/x(1/x +y/3)

= 1/x3+y3/27 + y/x2 + y2/3x

Hence, (1/x+y/3)3 = 1/x3+y3/27 + y/x2 + y2/3x

(b) (3/x-2/x2)

Given, (3/x – 2/x2)3                                                               [(a-b)3=a3-b3-3ab(a-b)]

we know that here, a = 3/x, b= y/3

By using formula:

(3/x – 2/x2)3 = (3/x)3 – (2/x2)3 -3(3/x)(2/x2)(3/x -2/x2)

= 27/x3 – 8/x6 – 3(6/x3)(3/x – 2/x2)

= 27/x3 – 8/x6 – 18/x3 (3/x – 2/x2)

= 27/x3 – 8/x6 – 54/x4 – 36/x5

Hence, (3/x – 2/x2)3 = 27/x3 – 8/x6 – 54/x4 – 36/x5

(c) (2x+3/x)

Given, (2x+3/x)3                                                               [(a+b)3=a3+b3+3ab(a+b)]

we know that, a = 2x, b = 3/x

By using formula,

(2x+3/x)3 = (2x)3 + (3/x)3 +3(2x)(3/x)(2x+3/x)

= 8x3 + 27/x3 + 18(2x+3/x)

= 8x3 + 27/x3 + 36x + 54/x

Hence, (2x+3/x)2 = 8x3 + 27/x3 + 36x + 54/x

(d) (4-1/3x)

Given, (4-1/3x)3

we know that, a = 4, b = 1/3x                                                               [(a-b)3=a3-b3-3ab(a-b)]

By using formula,

(4-1/3x)3 = 43-(1/3x)3 -3(4)(1/3x)(4-1/3x)

= 64 – (1/27x3) – 4/x(4 – 1/3x)

= 64 – (1/27x3) – 16/x + 4/3x

Hence, (4-1/3x) = 64 – (1/27x3) – 16/x + 4/3x

### Question 2. Simplify each of the following

(a) (x+3)3 + (x-3)3

(b) (x/2+y/3)3 – (x/2-y/3)3

(c) (x+2/x)3 + (x – 2/x)3

(d) (2x-5y)3 – (2x+5y)3

(a) (x+3)3 + (x-3)3

The above equation is in the form of a3+b3=(a+b)(a2+b2-ab)

we know that, a = (x+3), b= (x-3)

By using (a3+b3) formula

= (x+3+x-3)[(x+3)3+(x-3)3-(x+3)(x-3)]

= 2x[(x2+32+2*3*x)+(x2+32-2*3*x)-(x2-32)]

= 2x[(x2+9+6x)+(x2+9-6x)-x2+9]

= 2x[x2+27]

= 2x3+54x

Hence, the result of (x+3)3+(x-3)3 is 2x3+54x

(b) (x/2+y/3)3 – (x/2-y/3)3

The above equation is in the form of a3-b3=(a-b)(a2+b2+ab)

we know that, a = (x/2+y/3), b= (x/2-y/3)

By using (a3-b3) formula

= [(x/2+y/3)-(x/2-y/3)][(x/2+y/3)2+(x/2-y/3)2+(x/2+y/3)(x/2-y/3)]

= [2y/3][((x/2)2+(y/3)2+2(x/2)(y/3))+((x/2)2+(y/3)2 -2(x/2)(y/3)+x2/4-y2/9]

= (2y/3)[3x2/4+y2/9]

= x2y/2 +2y3/27

Hence, (x/2+y/3)3 – (x/2-y/3)3 = x2y/2 +2y3/27

(c) (x+2/x)3+(x-2/x)3

The above equation is in the form of a3+b3=(a+b)(a2+b2-ab)

we know that, a = (x+2/x), b= (x-2/x)

By using (a3+b3) formula

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

= (2x)[((x)2+(2/x)2+2(x)(2/x))+((x)2+(2/x)2-2(x)(2/x)) -(x2-(2/x)2)]

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

= (2x)[x2 +12/x2]

= 2x3+ 24/x

Hence, (x+2/x)3+(x-2/x)3 = 2x3+ 24/x

(d) (2x-5y)3-(2x+5y)3

The above equation is in the form of a3-b3=(a-b)(a2+b2+ab)

we know that, a = (2x-5y), b= (2x+5y)

By using (a3-b3) formula

= [(2x-5y)-(2x+5y)][(2x-5y)2+(2x+5y)2+(2x-5y)(2x+5y)]

= (-10y)[((2x)2+(5y)2-2(2x)(5y))+((2x)2+(5y)2+2(2x)(5y))+4x2-25y2]

= (-10y)[3(4x2)+25y2]

=(-10y)[12x2+25y2]

= -120x2y – 250y3

Hence, (2x-5y)3-(2x+5y)3 = -120x2y – 250y3

### Question 3. If a+b=10 and ab =21, Find the value of a3+b3 .

Solution:

Given,

a+b = 10, ab = 21

we know that, (a+b)3 = a3+b3+3ab(a+b) ———– 1

substitute a+b = 10, ab = 21 in eq -1

â‡’ (10)3 = a3+b3+3(21)(10)

â‡’ 1000 = a3+b3+630

â‡’ 1000 – 630 = a3+b3

â‡’ 370 = a3+b3

Hence, a3+b3=370

### Question 4. If a-b=4 and ab = 21, find the value of a3-b3.

Solution:

Given,

a-b = 4, ab = 21

we know that, (a-b)3=a3-b3 -3ab(a-b)

Substitute a-b=4, ab= 21 in eq-1

â‡’ (4)3 = a3-b3-3(21)(4)

â‡’ 64 = a3 -b3-252

â‡’ 64+252 = a3 – b3

â‡’ 316 = a3-b3

Hence, a3-b3 = 316

### Question 5. If (x+1/x) = 5, Find the value of x3+1/x3

Solution:

Given, (x+1/x) = 5

we know that, (a+b)3 = a3+b3+3ab(a+b) ————- 1

Substitute (x+1/x) = 5 in eq–1

(x+1/x)3=x3+(1/x)3+3(x)(1/x)(x+1/x)

(5)3 = x3+(1/x)3 + 3(5)

125 -15 = x3+(1/x)3

110 = x3+(1/x)3

Hence, the result is x3+1/x3 = 110

### Question 6. If (x-1/x) = 7, Find the value of x3-1/x3

Solution:

Given, If (x-1/x) = 7

we know that, (a-b)3 = a3-b3-3ab(a-b) ————- 1

substitute (x-1/x) = 7 in eq–1

(x-1/x)3 = x3 – 1/x3 -3(x)(1/x)(x-1/x)

(7)3 = x3-1/x3 -3(7)

343+21 = x3 – 1/x3

364 = x3 – 1/x3

Hence, the result is x3-1/x3 = 364

### Question 7. If (x-1/x) = 5, Find the value of x3-1/x3

Solution:

Given, If (x-1/x) = 5

we know that, (a-b)3 = a3-b3-3ab(a-b) ————- 1

Substitute (x-1/x) = 5 in eq–1

(x-1/x)3 = x3 – 1/x3 -3(x)(1/x)(x-1/x)

(5)3 = x3-1/x3 -3(5)

125+15 = x3 – 1/x3

x3 – 1/x3=140

Hence, the result is x3-1/x3 = 140

### Question 8. If (x2+1/x2) = 51, Find the value of x3-1/x3.

Solution:

Given, If (x2+1/x2) = 51

we know that, (a-b)2= a2+b2-2ab ————- 1

substitute (x2+1/x2) = 51 in eq–1

(x-1/x)2 = x2 +1/x2 -2(x)(1/x)

(x-1/x)2 = 51 -2

= 49

x-1/x = Â±7

we need to find x3-1/x3

So, a3-b3 = (a-b)(a2+b2+ab)

x3-1/x3 = (x-1/x)(x2+1/x2+x*1/x)

Substitute all known values here,

= 7(51+1)

=7(52)

x3-1/x3 = 364

Hence, the result is x3-1/x3 = 364

### Question 9. If (x2+1/x2) = 98, Find the value of x3+1/x3

Solution:

Given, (x2+1/x2) = 98

we know that, (x+y)2 = x2+y2+2xy ——– 1

substitute (x2+1/x2) = 98

(x+1/x)2=x2+1/x2+2(x)(1/x)

= 98 + 2

= 100

(x+1/x) = Â±10

We need to find x3+1/x3

(x+1/x) = 10 and (x2+1/x2) = 98

x3+1/x3 = 10(98-1)

= 970

Hence, the value of x3+1/x3 = 970

### Question 10. If 2x+3y=13 and xy=6, Find the value of 8x3+27y3

Solution:

Given, 2x+3y=13, xy=6

We know that,

â‡’ (2x+3y)3=133

â‡’ 8x3+27y3+3(2x)(3y)(2x+3y) = 2197

â‡’ 8x3+27y3+18xy(2x+3y) = 2197

â‡’ 8x3+27y3+18xy(2x+3y)= 2197

Substitute the values of 2x+3y=13, xy=6

â‡’ 8x3+27y3+18(6)(13)= 2197

â‡’ 8x3+27y3 = 2197 – 1404

â‡’ 8x3+27y3 = 793

Hence, the value of 8x3+27y3 = 793

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