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Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.3 | Set 2
• Last Updated : 03 Mar, 2021

### Question 11. If 3x-2y=11 and xy = 12, find the value of 27x3-8y3

Solution:

Given, 3x-2y=11 and xy=12

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

(3x-2y)3 = 113

27x3-8y3-3(12)(11)=1331

27x3-8y3=1331+2376

27x3-8y3=3707

Hence, the value of 27x3-8y3 = 3707

### Question 12. If (x4+1/x4)=119, Find the value of (x3-1/x3)

Solution:

Given, (x4+1/x4) =119 —— 1

we know that (x+y)2 = x2+y2+2xy

substitute the given value in eq-1

(x2+1/x2)2 = x4+1/x4+2(x2)(1/x2)

= x4+1/x4+2

= 119+2

= 121

(x2+1/x2)2 = ±11

Now, find (x-1/x)

we know that (x-y)2=x2+y2-2xy

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

= x2+1/x2 -2

= 11 -2

= 9

(x-1/x) = ±3

we need to find x3-1/x3

we know that , a3-b3=(a-b)(a2+b2-ab)

x3-(1/x)3=(x-1/x)(x2+(1/x)2-x*1/x)

Here, x2+1/x2 = 11 and x-1/x=3

x3-1/x3 = 3(11+1)

= 3(12)

= 36

Hence, the value of x3-1/x3=36

### Question 13. Evaluate each of the following:

(a) (103)3

(b) (98)3

(c) (9.9)3

(d) (10.4)3

(e) (598)3

(f) (99)3

Solution:

Given:

(a) (103)3

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

(103)3 can be written as (100+3)3

Here, a=100 and b=3

(103)3 = (100+3)3

=1003+33+3(100)(3)(100+3)

=1000000+27+900(103)

=1092727

The value of (103)3=1092727

(b) (98)3

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

(98)3 = (100-2)3

= 1003-23-3(100)(2)(100-2)

= 1000000-8-600(98)

= 941192

The value of (98)3 = 941192

(c) (9.9)3

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

(10-0.1)3 = (10)3-(0.1)3-3(10)(0.1)(10-0.1)

= 1000 – 0.001-3(9.9)

= 970.299

The value of (9.9)3=970.299

(d) (10.4)3

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

(10+0.4)3=(10)3+(0.4)3+3(10)(0.4)(10+0.4)

= 1000+0.064+12(10.4)

= 1124.864

The value of (10.4)3=1124.864

(e) (598)3

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

(600-2)3 = (600)3-23-3(600)(2)(600-2)

= 216000000 – 8 -(3600*598)

= 216000000 -8 – 2152800

= 213847192

The value of (598)3 = 213847192

(f) (99)3

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

(100-1)3 = (100)3 -13 -3(100)(1)(100-1)

= 1000000 – 1 -300*99

= 1000000 – 1 -29700

= 970299

The value of (99)3 = 970299

### Question 14. Evaluate each of the following

(a) 1113 – 893

(b) 463 +343

(c) 1043+963

(d) 933 – 1073

Solution:

Given:

(a) 1113 – 893

The above equation can be written as (100+11)3 – (100-11)3

we know that , (a+b)3 -(a-b)3= 2[b3+3a2b]

Here, a=100 b=11

(100+11)3 – (100-11)3 = 2[113+3(100)2(11)]

= 2[1331 + 330000]

= 2

= 662662

The value of 1113-893 = 662662

(b) 463 + 343

The above equation can be written as (40+6)3 – (40-6)3

we know that , (a+b)3 +(a-b)3= 2[a3+3ab2]

Here, a = 40 and b=6

(40+6)3 – (40-6)3 = 2[(40)3+3(6)2(40)]

= 2[64000+3*36*40]

=2

= 136640

The value of 463 +343=136640

(c) 1043 +963

The above equation can be written as (100+4)3 + (100-4)3

we know that, (a+b)3 +(a-b)3= 2[a3+3ab2]

here, a = 100 , b= 4

(100+4)3 +(100-4)3 = 2[(100)3 +3(100)(4)2]

= 2[1000000 + 300*16]

= 2

= 2009600

The value of 1043 + 963 = 2009600

(d) 933 – 1073

The above equation can be written as (100-7)3 – (100+7)3

we know that, (a-b)3 -(a+b)3= -2[b3+3a2b]

here, a = 100 , b= 7

(100-7)3 – (100+7)3 = -2[73+3*(100)2*7]

= -2

= -420686

The value of 933 – 1073 = -420686

### Question 15. If x+1/x = 3, calculate x2+1/x2, x3+1/x3, x4+1/x4

Solution:

Given, x+1/x=3

we know that (x+y)2= x2+y2+2xy

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

(3)2 = x2+(1/x)2 +2

x2+1/x2 =7

squaring on both the sid

(x2+1/x2)2 = 49

x4+1/x4+2(x2)(1/x2) = 49

x4 +1/x4 = 49-2

x4+1/x4 = 47

again cubing on both the sides ,

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

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

x3+1/x3=27-9

x3+1/x3 = 18

The value x2+1/x2 =7, x3+1/x3 = 18, x4+1/x4 = 47

### Question 16. If x4+1/x4=194, calculate x2+1/x2, x3+1/x3,  x+1/x

Solution:

Given,

x4+1/x4=194 —– 1

add and substract (2*x2*1/x2) on the left side in above given equation

x4+1/x4 +2*x2*1/x2 -2*x*1/x2 = 194

x4+1/x4+2*x2*1/x2 -2 =194

(x2)2+(1/x2)2+ 2*x2*1/x2 = 196

(x2+1/x2)2 = 196

(x2+1/x2) = 14 ——— 2

add and subtract (2*x*1/x) on the left side in above given equation

x2+1/x2+2*x*1/x-2*x*1/x =14

(x+1/x)2 = 14 +2

(x+1/x) = 4 ———— 3

Now cubing eq-3 on both sides.

(x+1/x)3 = 43

x3+1/x3+3*x*1/x(x+1/x) = 64

x3+1/x3 +3*4 = 64

x3+1/x3 = 64 -12

= 52

Hence, the values of (x2+1/x2) = 14, (x3+1/x3) = 52, (x+1/x) = 4

### Question 17. Find the value of 27x3+8y3, if

(a) 3x+2y=14 and xy = 8

(b) 3x+2y = 20 and xy=14/9

Solution:

(a) Given, 3x+2y = 14 and xy = 8

cubing on both the sides

(3x+2y)3 = 143

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

27x3+8y3+3(3x)(2y)(3x+2y) = 2744

27x3+8y3+18xy(3x+2y) = 2744

27x3+8y3+18*8*14 = 2744

27x3+8y3 = 2744 – 2016

27x3 +8y3 = 728

Hence, the value of 27x3 +8y3 = 728

(b) Given, 3x+2y = 20 and xy=14/9

cubing on both the sides

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

27x3+8y3+3(3x)(2y)(3x+2y) = 8000

27x3+8y3+18xy(3x+2y) = 8000

27x3+8y3+18*14/9*20 = 8000

27x3+8y3 = 8000 – 560

= 7440

Hence, the value of 27x3+8y3 = 7440

### Question 18. Find the value of 64x3-125z3, if 4x-5z=16 and xz=12

Solution:

Given, 64x3 – 125z3

Here, 4x -5z = 16 and xz = 12

cubing (4x-5z)3 = 163

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

(4x -5z)3 = (4x)3-(5z)3-3(4x)(5z)(4x-5z)

(16)3 = 64x3 -125z3-60(4x-5z)

4096 = 64x3-125z3-60(16)

64x3-125z3 = 4096 + 960

= 5056

Hence, the value of 64x3 – 125z3 = 5056

### Question 19. If x-1/x = , find the value of x3-1/x3

Solution:

Given, x-1/x = cubing both the sides,

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

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

(3+2\sqrt{2})^3 = x3-1/x3 -3(3 )  x3-1/x3 = 108+ Hence, the value of x3-1/x3 = 108+ Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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