RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

In this chapter, we provide RD Sharma Solutions for Chapter 4 Algebraic Identities Ex 4.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 4 Algebraic Identities Ex 4.3 Maths pdf, free RD Sharma Solutions for Chapter 4 Algebraic Identities Ex 4.3 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Solutions for Class 9 Chapter 4 Algebraic Identities Ex 4.3 Download PDF

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

Solution:

Question 2.
If a + b = 10 and ab = 21, find the value of a3 + b3.
Solution:
a + b = 10, ab = 21
Cubing both sides,
(a + b)3 = (10)3
⇒ a3 + 63 + 3ab (a + b) = 1000
⇒  a3 + b3 + 3 x 21 x 10 = 1000
⇒  a3 + b3 + 630 = 1000
⇒  a3 + b3 = 1000 – 630 = 370
∴ a3 + b3 = 370

Question 3.
If a – b = 4 and ab = 21, find the value of a3-b3.
Solution:
a – b = 4, ab= 21
Cubing both sides,
⇒ (a – A)3 = (4)3
⇒ a3 – b3 – 3ab (a – b) = 64
⇒ a3-i3-3×21 x4 = 64
⇒  a3 – 63 – 252 = 64
⇒  a3 – 63 = 64 + 252 =316
∴ a3 – b3 = 316

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3.Solution:
2x + 3y = 13, xy = 6
Cubing both sides,
(2x + 3y)3 = (13)3
⇒ (2x)3 + (3y)3 + 3 x 2x x 3X2x + 3y) = 2197
⇒ 8x3 + 27y3 + 18xy(2x + 3y) = 2197
⇒ 8x3 + 27y3 + 18 x 6 x 13 = 2197
⇒ 8X3 + 27y3 + 1404 = 2197
⇒  8x3 + 27y3 = 2197 – 1404 = 793
∴ 8x3 + 27y3 = 793

Question 10.
If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3.
Solution:
3x – 2y = 11 and xy = 12 Cubing both sides,
(3x – 2y)3 = (11)3
⇒  (3x)3 – (2y)3 – 3 x 3x x 2y(3x – 2y) =1331
⇒  27x3 – 8y3 – 18xy(3x -2y) =1331
⇒   27x3 – 8y3 – 18 x 12 x 11 = 1331
⇒  27x3 – 8y3 – 2376 = 1331
⇒  27X3 – 8y3 = 1331 + 2376 = 3707
∴ 2x3 – 8y3 = 3707

Question 11.
Evaluate each of the following:
(i)  (103)3
(ii) (98)3
(iii) (9.9)3
(iv) (10.4)3
(v) (598)3
(vi) (99)3
Solution:
We know that (a + bf = a3 + b3 + 3ab(a + b) and (a – b)3= a3 – b3 – 3 ab(a – b)
Therefore,
(i)  (103)3 = (100 + 3)3
= (100)3 + (3)3 + 3 x 100 x 3(100 + 3)    {∵ (a + b)3 = a3 + b3 + 3ab(a + b)}
= 1000000 + 27 + 900 x 103
= 1000000 + 27 + 92700
= 1092727
(ii) (98)3 = (100 – 2)3
= (100)3 – (2)3 – 3 x 100 x 2(100 – 2)
= 1000000 – 8 – 600 x 98
= 1000000 – 8 – 58800
= 1000000-58808
= 941192
(iii) (9.9)3 = (10 – 0.1)3
= (10)3 – (0.1)3 – 3 X 10 X 0.1(10 – 0.1)
= 1000 – 0.001 – 3 x 9.9
= 1000 – 0.001 – 29.7
= 1000 – 29.701
= 970.299
(iv) (10.4)3 = (10 + 0.4)3
= (10)3 + (0.4)3 + 3 x 10 x 0.4(10 + 0.4)
= 1000 + 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 = 1124.864
(v) (598)3 = (600 – 2)3
= (600)3 – (2)3 – 3 x 600 x 2 x (600 – 2)
= 216000000 – 8 – 3600 x 598
= 216000000 – 8 – 2152800
= 216000000 – 2152808
= 213847192
(vi) (99)3 = (100 – 1)3
= (100)3 – (1)3 – 3 x 100 x 1 x (100 – 1)
= 1000000 – 1 – 300 x 99
= 1000000 – 1 – 29700
= 1000000 – 29701
= 970299

Question 12.
Evaluate each of the following:
(i)  1113 – 893
(ii) 463 + 343
(iii) 1043 + 963
(iv) 933 – 1073
Solution:
We know that a3 + b3 = (a + bf – 3ab(a + b) and a3 – b3 = (a – bf + 3 ab(a – b)
(i) 1113 – 893
= (111 – 89)3 + 3 x ill x 89(111 – 89)
= (22)3 + 3 x 111 x 89 x 22
= 10648 + 652014 = 662662
(OR)
(a + b)3 – (a – b)3 = 2(b3 + 3a2b)
= 1113 – 893 = (100 + 11)3 – (100 – 11)3
= 2(113 + 3 x 1002 x 11]
= 2(1331 + 330000]
= 331331 x 2 = 662662
(a + b)3 + (a- b)3 = 2(b3 + 3ab2)
(ii) 463 + 343 = (40 + 6)3 + (40 – 6)3
= 2[(40)3 + 3 x 40 x 62]
= 2[64000 + 3 x 40 x 36]
= 2[64000 + 4320]
= 2 x 68320 = 136640
(iii) 1043 + 963 = (100 + 4)3 + (100 – 96)3
= 2 [a3 + 3 ab2]
= 2[(100)3 + 3 x 100 x (4)2]
= 2[ 1000000 + 300 x 16]
= 2[ 1000000 + 4800]
= 1004800 x 2 = 2009600
(iv) 933 – 1073 = -[(107)3 – (93)3]
= -[(100 + If – (100 – 7)3]
= -2[b3 + 3a2b)]
= -2[(7)3 + 3(100)2 x 7]
= -2(343 + 3 x 10000 x 7]
= -2[343 + 210000]
= -2[210343] = -420686

Question 13.

Solution:

Question 14.
Find the value of 27X3 + 8y3 if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy = 149
Solution:

Question 15.
Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz = 12.
Solution:
4x – 5z = 16, xz = 12
Cubing both sides,
(4x – 5z)3 = (16)3
⇒ (4x)3 – (5y)3 – 3 x 4x x 5z(4x – 5z) = 4096
⇒ 64x3 – 125z3 – 3 x 4 x 5 x xz(4x – 5z) = 4096
⇒  64x3 – 125z3 – 60 x 12 x 16 = 4096
⇒ 64x3 – 125z3 – 11520 = 4096
⇒  64x3 – 125z3 = 4096 + 11520 = 15616

Question 16.

Solution:

Question 17.
Simplify each of the following:

Solution:

Question 18.

Solution:

Question 19.

Solution:

All Chapter RD Sharma Solutions For Class 9 Maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class 9

All Subject NCERT Solutions For Class 9

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.