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RD Sharma Solutions Class 9 Maths Chapter 4 Algebraic Identities

Mathematics holds significant importance for students in CBSE Class 9, shaping their academic journey. To ensure students are well-prepared, we provide comprehensive support. In this article, students can access RD Sharma Class 9 Solutions for Maths Chapter 4.
authorImageAnanya Gupta6 Apr, 2024
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RD Sharma Solutions Class 9 Maths Chapter 4

RD Sharma Solutions Class 9 Maths Chapter 4: In RD Sharma Class 9 Maths Chapter 4, you'll find solutions for algebraic identities. These solutions help explain the problems in the textbook. They make it easier to understand algebraic equations that are always true, no matter the values of the variables. By using these solutions, you can better understand algebraic concepts and how to solve related math problems. They're designed to be easy to follow and help you prepare well for exams.

By using RD Sharma Class 9 Solutions, students can practice important questions for their final exams, improve time management, and enhance logical thinking skills. These solutions are created by experts at Physics Wallah to help students understand concepts easily. They are available in PDF format for easy access both online and offline.

CBSE Class 9 Science Syllabus 2024-25

RD Sharma Solutions Class 9 Maths Chapter 4 Algebraic Identities PDF

Here we have provided RD Sharma Class 9 Solutions Maths Chapter 4 solutions for the students to help them ace their examinations. Students can refer to these solutions and practice these questions to score better in the exams.

RD Sharma Solutions Class 9 Maths Chapter 4 PDF

RD Sharma Solutions Class 9 Maths Chapter 4 Algebraic Identities

RD Sharma Solutions Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.1 Page No: 4.6

Question 1: Evaluate each of the following using identities:

(i) (2x – 1/x) 2

(ii) (2x + y) (2x – y)

(iii) (a 2 b – b 2 a) 2

(iv) (a – 0.1) (a + 0.1)

(v) (1.5.x 2 – 0.3y 2 ) (1.5x 2 + 0.3y 2 )

Solution :

(i) (2x – 1/x) 2

[Use identity: (a – b) 2 = a 2 + b 2 – 2ab ] (2x – 1/x) 2 = (2x) 2 + (1/x) 2 – 2 (2x)(1/x) = 4x 2 + 1/x 2 – 4

(ii) (2x + y) (2x – y)

[Use identity: (a – b)(a + b) = a 2 – b 2 ] (2x + y) (2x – y) = (2x ) 2 – (y) 2 = 4x 2 – y 2

(iii) (a 2 b – b 2 a) 2

[Use identity: (a – b) 2 = a 2 + b 2 – 2ab ] (a 2 b – b 2 a) 2 = (a 2 b) 2 + (b 2 a) 2 – 2 (a 2 b)( b 2 a) = a 4 b 2 + b 4 a 2 – 2 a 3 b 3

(iv) (a – 0.1) (a + 0.1)

[Use identity: (a – b)(a + b) = a 2 – b 2 ] (a – 0.1) (a + 0.1) = (a) 2 – (0.1) 2 = (a) 2 – 0.01

(v) (1.5 x 2 – 0.3y 2 ) (1.5 x 2 + 0.3y 2 )

[Use identity: (a – b)(a + b) = a 2 – b 2 ] (1.5 x 2 – 0.3y 2 ) (1.5x 2 + 0.3y 2 ) = (1.5 x 2 ) 2 – (0.3y 2 ) 2 = 2.25 x 4 – 0.09y 4

CBSE Class 10 Result 2024 Expected To Be Out Soon

Question 2: Evaluate each of the following using identities:

(i) (399) 2

(ii) (0.98) 2

(iii) 991 x 1009

(iv) 117 x 83

Solution :

(i)

RD sharma class 9 maths chapter 4 ex 4.1 question 2 part 1 Solution

(ii)

RD sharma class 9 maths chapter 4 ex 4.1 question 2 part 2 Solution

(iii)

RD sharma class 9 maths chapter 4 ex 4.1 question 2 part 3 Solution

(iv)

RD sharma class 9 maths chapter 4 ex 4.1 question 2 part 4 Solution

Question 3: Simplify each of the following:

(i) 175 x 175 +2 x 175 x 25 + 25 x 25

(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22

(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24

(iv)

RD sharma class 9 maths chapter 4 ex 4.1 question 3

Solution :

(i) 175 x 175 +2 x 175 x 25 + 25 x 25 = (175) 2 + 2 (175) (25) + (25) 2

= (175 + 25) 2 [Because a 2 + b 2 +2ab = (a+b) 2 ] = (200) 2 = 40000 So, 175 x 175 +2 x 175 x 25 + 25 x 25 = 40000.

(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22

= (322) 2 – 2 x 322 x 22 + (22) 2 = (322 – 22) 2 [Because a 2 + b 2 -2ab = (a-b) 2 ] = (300) 2 = 90000 So, 322 x 322 – 2 x 322 x 22 + 22 x 22= 90000.

(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24

= (0.76) 2 + 2 x 0.76 x 0.24 + (0.24) 2 = (0.76+0.24) 2 [ Because a 2 + b 2 +2ab = (a+b) 2 ] = (1.00) 2 = 1 So, 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24 = 1.

(iv)

RD sharma class 9 maths chapter 4 ex 4.1 question 3 part 4 solution

Question 4: If x + 1/x = 11, find the value of x 2 +1/x 2 .

Solution:

RD sharma class 9 maths chapter 4 ex 4.1 question 4

Question 5: If x – 1/x = -1, find the value of x 2 +1/x 2 .

Solution:

RD sharma class 9 maths chapter 4 ex 4.1 question 5 solution

Chapter-wise Solutions for RD Sharma Class 9
RD Sharma Class 9 Solutions Maths
RD Sharma Class 9 Solutions Maths Chapter 1
RD Sharma Class 9 Solutions Maths Chapter 2
RD Sharma Class 9 Solutions Maths Chapter 3
RD Sharma Class 9 Solutions Maths Chapter 5
RD Sharma Class 9 Solutions Maths Chapter 6
RD Sharma Class 9 Solutions Maths Chapter 7
RD Sharma Class 9 Solutions Maths Chapter 8
RD Sharma Class 9 Solutions Maths Chapter 9
RD Sharma Class 9 Solutions Maths Chapter 10
RD Sharma Class 9 Solutions Maths Chapter 11
RD Sharma Class 9 Solutions Maths Chapter 12
RD Sharma Class 9 Solutions Maths Chapter 13
RD Sharma Class 9 Solutions Maths Chapter 14
RD Sharma Class 9 Solutions Maths Chapter 15
RD Sharma Class 9 Solutions Maths Chapter 16
RD Sharma Class 9 Solutions Maths Chapter 17
RD Sharma Class 9 Solutions Maths Chapter 18
RD Sharma Class 9 Solutions Maths Chapter 19
RD Sharma Class 9 Solutions Maths Chapter 20
RD Sharma Class 9 Solutions Maths Chapter 21
RD Sharma Class 9 Solutions Maths Chapter 22
RD Sharma Class 9 Solutions Maths Chapter 23
RD Sharma Class 9 Solutions Maths Chapter 24
RD Sharma Class 9 Solutions Maths Chapter 25

RD Sharma Solutions Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.2 Page No: 4.11

Question 1: Write the following in the expanded form:

(i) (a + 2b + c) 2

(ii) (2a − 3b − c) 2

(iii) (−3x+y+z) 2

(iv) (m+2n−5p) 2

(v) (2+x−2y) 2

(vi) (a 2 +b 2 +c 2 ) 2

(vii) (ab+bc+ca) 2

(viii) (x/y+y/z+z/x)2

(ix) (a/bc + b/ac + c/ab) 2

(x) (x+2y+4z) 2

(xi) (2x−y+z) 2

(xii) (−2x+3y+2z) 2

Solution :

Using identities: (x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2xz

(i) (a + 2b + c) 2

= a 2 + (2b) 2 + c 2 + 2a(2b) + 2ac + 2(2b)c = a 2 + 4b 2 + c 2 + 4ab + 2ac + 4bc

(ii) (2a − 3b − c) 2

= [(2a) + (−3b) + (−c)] 2 = (2a) 2 + (−3b) 2 + (−c) 2 + 2(2a)(−3b) + 2(−3b)(−c) + 2(2a)(−c) = 4a 2 + 9b 2 + c 2 − 12ab + 6bc − 4ca

(iii) (−3x+y+z) 2

= [(−3x) 2 + y 2 + z 2 + 2(−3x)y + 2yz + 2(−3x)z = 9x 2 + y 2 + z 2 − 6xy + 2yz − 6xz

(iv) (m+2n−5p) 2

= m 2 + (2n) 2 + (−5p) 2 + 2m × 2n + (2×2n×−5p) + 2m × −5p = m 2 + 4n 2 + 25p 2 + 4mn − 20np − 10pm

(v) (2+x−2y) 2

= 2 2 + x 2 + (−2y) 2 + 2(2)(x) + 2(x)(−2y) + 2(2)(−2y) = 4 + x 2 + 4y 2 + 4 x − 4xy − 8y

(vi) (a 2 +b 2 +c 2 ) 2

= (a 2 ) 2 + (b 2 ) 2 + (c 2 ) 2 + 2a 2 b 2 + 2b 2 c 2 + 2a 2 c 2 = a 4 + b 4 + c 4 + 2a 2 b 2 + 2b 2 c 2 + 2c 2 a 2

(vii) (ab+bc+ca) 2

= (ab) 2 + (bc) 2 + (ca) 2 + 2(ab)(bc) + 2(bc)(ca) + 2(ab)(ca) = a 2 b 2 + b 2 c 2 + c 2 a 2 + 2(ac)b 2 + 2(ab)(c) 2 + 2(bc)(a) 2

(viii) (x/y+y/z+z/x) 2

RD sharma class 9 maths chapter 4 ex 4.2 question 1 solution

(ix) (a/bc + b/ac + c/ab) 2

RD sharma class 9 maths chapter 4 ex 4.2 question 1 solution

(x) (x+2y+4z) 2

= x 2 + (2y) 2 + (4z) 2 + (2x)(2y) + 2(2y)(4z) + 2x(4z) = x 2 + 4y 2 + 16z 2 + 4xy + 16yz + 8xz

(xi) (2x−y+z) 2

= (2x) 2 + (−y) 2 + (z) 2 + 2(2x)(−y) + 2(−y)(z) + 2(2x)(z) = 4x 2 + y 2 + z 2 − 4xy−2yz+4xz

(xii) (−2x+3y+2z) 2

= (−2x) 2 + (3y) 2 + ( 2z) 2 + 2(−2x)(3y)+2(3y)(2z)+2(−2x)(2z) = 4x 2 + 9y 2 + 4z 2 −12xy+12yz−8xz

Question 2: Simplify

(i) (a + b + c) 2 + (a − b + c) 2

(ii) (a + b + c) 2 − (a − b + c) 2

(iii) (a + b + c) 2 + (a – b + c) 2 + (a + b − c) 2

(iv) (2x + p − c) 2 − (2x − p + c) 2

(v) (x 2 + y 2 − z 2 ) 2 − (x 2 − y 2 + z 2 ) 2

Solution:

(i) (a + b + c) 2 + (a − b + c) 2

= (a 2 + b 2 + c 2 + 2ab+2bc+2ca) + (a 2 + (−b) 2 + c 2 −2ab−2bc+2ca) = 2a 2 + 2 b 2 + 2c 2 + 4ca

(ii) (a + b + c) 2 − (a − b + c) 2

= (a 2 + b 2 + c 2 + 2ab+2bc+2ca) − (a 2 + (−b) 2 + c 2 −2ab−2bc+2ca) = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca − a 2 − b 2 − c 2 + 2ab + 2bc − 2ca = 4ab + 4bc

(iii) (a + b + c) 2 + (a – b + c) 2 + (a + b − c) 2

= a 2 + b 2 + c 2 + 2ab + 2bc + 2ca + (a 2 + b 2 + (c) 2 − 2ab − 2cb + 2ca) + (a 2 + b 2 + c 2 + 2ab − 2bc – 2ca) = 3 a 2 + 3b 2 + 3c 2 + 2ab − 2bc + 2ca

(iv) (2x + p − c) 2 − (2x − p + c) 2

= [4x 2 + p 2 + c 2 + 4xp − 2pc − 4xc] − [4x 2 + p 2 + c 2 − 4xp− 2pc + 4xc] = 4x 2 + p 2 + c 2 + 4xp − 2pc − 4cx − 4x 2 − p 2 − c 2 + 4xp + 2pc− 4cx = 8xp − 8xc = 8(xp − xc)

(v) (x 2 + y 2 − z 2 ) 2 − (x 2 − y 2 + z 2 ) 2

= (x 2 + y 2 + (−z) 2 ) 2 − (x 2 − y 2 + z 2 ) 2 = [x 4 + y 4 + z 4 + 2x 2 y 2 – 2y 2 z 2 – 2x 2 z 2 − [x 4 + y 4 + z 4 − 2x 2 y 2 − 2y 2 z 2 + 2x 2 z 2 ] = 4x 2 y 2 – 4z 2 x 2

Question 3: If a + b + c = 0 and a 2 + b 2 + c 2 = 16, find the value of ab + bc + ca.

Solution :

a + b + c = 0 and a 2 + b 2 + c 2 = 16 (given) Choose a + b + c = 0 Squaring both sides, (a + b + c) 2 = 0 a 2 + b 2 + c 2 + 2(ab + bc + ca) = 0 16 + 2(ab + bc + c) = 0 2(ab + bc + ca) = -16 ab + bc + ca = -16/2 = -8 or ab + bc + ca = -8

RD Sharma Solutions Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.3 Page No: 4.19

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

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

(ii) (3/x – 2/x 2 )

(iii) (2x + 3/x)

(iv) (4 – 1/3x)

Solution:

[Using identities: (a + b) 3 = a 3 + b 3 + 3ab(a + b) and (a – b) 3 = a 3 – b 3 – 3ab(a – b) ]

(i)

RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 1 solution

(ii)

RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 2 solution

(iii)

RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 3 solution

(iv)

RD sharma class 9 maths chapter 4 ex 4.3 Q1 part 4 solution

Question 2: Simplify each of the following:

(i) (x + 3) 3 + (x – 3) 3

(ii) (x/2 + y/3) 3 – (x/2 – y/3) 3

(iii) (x + 2/x) 3 + (x – 2/x) 3

(iv) (2x – 5y) 3 – (2x + 5y) 3

Solution:

[Using identities: a 3 + b 3 = (a + b)(a 2 + b 2 – ab) a 3 – b 3 = (a – b)(a 2 + b 2 + ab) (a + b)(a-b) = a 2 – b 2 (a + b) 2 = a 2 + b 2 + 2ab and (a – b) 2 = a 2 + b 2 – 2ab]

(i) (x + 3) 3 + (x – 3) 3

Here a = (x + 3), b = (x – 3) RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 1

(ii) (x/2 + y/3) 3 – (x/2 – y/3) 3

Here a = (x/2 + y/3) and b = (x/2 – y/3) RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 2

(iii) (x + 2/x) 3 + (x – 2/x) 3

Here a = (x + 2/x) and b = (x – 2/x) RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 3

(iv) (2x – 5y) 3 – (2x + 5y) 3

Here a = (2x – 5y) and b = 2x + 5y RD sharma class 9 maths chapter 4 ex 4.3 Q2 solution part 4

Question 3: If a + b = 10 and ab = 21, find the value of a 3 + b 3 .

Solution :

a + b = 10, ab = 21 (given) Choose a + b = 10 Cubing both sides, (a + b) 3 = (10) 3 a 3 + b 3 + 3ab(a + b) = 1000 a 3 + b 3 + 3 x 21 x 10 = 1000 (using given values) a 3 + b 3 + 630 = 1000 a 3 + b 3 = 1000 – 630 = 370 or a 3 + b 3 = 370

Question 4: If a – b = 4 and ab = 21, find the value of a 3 – b 3 .

Solution :

a – b = 4, ab= 21 (given) Choose a – b = 4 Cubing both sides, (a – b) 3 = (4) 3 a 3 – b 3 – 3ab (a – b) = 64 a 3 – b 3 – 3 × 21 x 4 = 64 (using given values) a 3 – b 3 – 252 = 64 a 3 – b 3 = 64 + 252 = 316 Or a 3 – b 3 = 316

Question 5: If x + 1/x = 5, find the value of x 3 + 1/x 3 .

Solution :

Given: x + 1/x = 5 Apply Cube on x + 1/x RD sharma class 9 maths chapter 4 ex 4.3 Q5 solution

Question 6: If x – 1/x = 7, find the value of x 3 – 1/x 3 .

Solution :

Given: x – 1/x = 7 Apply Cube on x – 1/x RD sharma class 9 maths chapter 4 ex 4.3 Q6 solution

Question 7: If x – 1/x = 5, find the value of x 3 – 1/x 3 .

Solution:

Given: x – 1/x = 5 Apply Cube on x – 1/x

RD sharma class 9 maths chapter 4 ex 4.3 Ques 7 solution

Question 8: If (x 2 + 1/x 2 ) = 51, find the value of x 3 – 1/x 3 .

Solution:

We know that: (x – y) 2 = x 2 + y 2 – 2xy Replace y with 1/x, we get (x – 1/x) 2 = x 2 + 1/x 2 – 2 Since (x 2 + 1/x 2 ) = 51 (given) (x – 1/x) 2 = 51 – 2 = 49 or (x – 1/x) = ±7 Now, Find x 3 – 1/x 3 We know that, x 3 – y 3 = (x – y)(x 2 + y 2 + xy) Replace y with 1/x, we get x 3 – 1/x 3 = (x – 1/x)(x 2 + 1/x 2 + 1) Use (x – 1/x) = 7 and (x 2 + 1/x 2 ) = 51 x 3 – 1/x 3 = 7 x 52 = 364 x 3 – 1/x 3 = 364

Question 9: If (x 2 + 1/x 2 ) = 98, find the value of x 3 + 1/x 3 .

Solution:

We know that: (x + y) 2 = x 2 + y 2 + 2xy Replace y with 1/x, we get (x + 1/x) 2 = x 2 + 1/x 2 + 2 Since (x 2 + 1/x 2 ) = 98 (given) (x + 1/x) 2 = 98 + 2 = 100 or (x + 1/x) = ±10 Now, Find x 3 + 1/x 3 We know that, x 3 + y 3 = (x + y)(x 2 + y 2 – xy) Replace y with 1/x, we get x 3 + 1/x 3 = (x + 1/x)(x 2 + 1/x 2 – 1) Use (x + 1/x) = 10 and (x 2 + 1/x 2 ) = 98 x 3 + 1/x 3 = 10 x 97 = 970 x 3 + 1/x 3 = 970

Question 10: If 2x + 3y = 13 and xy = 6, find the value of 8x 3 + 27y 3 .

Solution:

Given: 2x + 3y = 13, xy = 6 Cubing 2x + 3y = 13 both sides, we get (2x + 3y) 3 = (13) 3 (2x) 3 + (3y) 3 + 3( 2x )(3y) (2x + 3y) = 2197 8x 3 + 27y 3 + 18xy(2x + 3y) = 2197 8x 3 + 27y 3 + 18 x 6 x 13 = 2197 8x 3 + 27y 3 + 1404 = 2197 8x 3 + 27y 3 = 2197 – 1404 = 793 8x 3 + 27y 3 = 793

Question 11: If 3x – 2y= 11 and xy = 12, find the value of 27x 3 – 8y 3 .

Solution:

Given: 3x – 2y = 11 and xy = 12 Cubing 3x – 2y = 11 both sides, we get (3x – 2y) 3 = (11) 3 (3x) 3 – (2y) 3 – 3 ( 3x)( 2y) (3x – 2y) =1331 27x 3 – 8y 3 – 18xy(3x -2y) =1331 27x 3 – 8y 3 – 18 x 12 x 11 = 1331 27x 3 – 8y 3 – 2376 = 1331 27x 3 – 8y 3 = 1331 + 2376 = 3707 27x 3 – 8y 3 = 3707

RD Sharma Solutions Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.4 Page No: 4.23

Question 1: Find the following products:

(i) (3x + 2y)(9x 2 – 6xy + 4y 2 )

(ii) (4x – 5y)(16x 2 + 20xy + 25y 2 )

(iii) (7p 4 + q)(49p 8 – 7p 4 q + q 2 )

(iv) (x/2 + 2y)(x 2 /4 – xy + 4y 2 )

(v) (3/x – 5/y)(9/x 2 + 25/y 2 + 15/xy)

(vi) (3 + 5/x)(9 – 15/x + 25/x 2 )

(vii) (2/x + 3x)(4/x 2 + 9x 2 – 6)

(viii) (3/x – 2x 2 )(9/x 2 + 4x 4 – 6x)

(ix) (1 – x)(1 + x + x 2 )

(x) (1 + x)(1 – x + x 2 )

(xi) (x 2 – 1)(x 4 + x 2 +1)

(xii) (x 3 + 1)(x 6 – x 3 + 1)

Solution:

(i) (3x + 2y)(9x 2 – 6xy + 4y 2 )

= (3x + 2y)[(3x) 2 – (3x)(2y) + (2y) 2 )] We know, a 3 + b 3 = (a + b)(a 2 + b 2 – ab) = (3x) 3 + (2y) 3 = 27x 3 + 8y 3

(ii) (4x – 5y)(16x 2 + 20xy + 25y 2 )

= (4x – 5y)[(4x) 2 + (4x)(5y) + (5y) 2 )] We know, a 3 – b 3 = (a – b)(a 2 + b 2 + ab) = (4x) 3 – (5y) 3 = 64x 3 – 125y 3

(iii) (7p 4 + q)(49p 8 – 7p 4 q + q 2 )

= (7p 4 + q)[(7p 4 ) 2 – (7p 4 )(q) + (q) 2 )] We know, a 3 + b 3 = (a + b)(a 2 + b 2 – ab) = (7p 4 ) 3 + (q) 3 = 343 p 12 + q 3

(iv) (x/2 + 2y)(x 2 /4 – xy + 4y 2 )

We know, a 3 – b 3 = (a – b)(a 2 + b 2 + ab) (x/2 + 2y)(x 2 /4 – xy + 4y 2 ) RD sharma class 9 maths chapter 4 ex 4.4 question 2 Solution

(v) (3/x – 5/y)(9/x 2 + 25/y 2 + 15/xy)

RD sharma class 9 maths chapter 4 ex 4.4 [Using a 3 – b 3 = (a – b)(a 2 + b 2 + ab) ]

(vi) (3 + 5/x)(9 – 15/x + 25/x 2 )

RD sharma class 9 maths chapter 4 ex 4.4 solution [Using: a 3 + b 3 = (a + b)(a 2 + b 2 – ab)]

(vii) (2/x + 3x)(4/x 2 + 9x 2 – 6)

RD sharma class 9 maths chapter 4 ex 4.4 question 1 [Using: a 3 + b 3 = (a + b)(a 2 + b 2 – ab)]

(viii) (3/x – 2x 2 )(9/x 2 + 4x 4 – 6x)

RD sharma class 9 maths chapter 4 ex 4.4 question 1 solution [Using : a 3 – b 3 = (a – b)(a 2 + b 2 + ab)]

(ix) (1 – x)(1 + x + x 2 )

And we know, a 3 – b 3 = (a – b)(a 2 + b 2 + ab) (1 – x)(1 + x + x 2 ) can be written as (1 – x)[(1 2 + (1)(x)+ x 2 )] = (1) 3 – (x) 3 = 1 – x 3

(x) (1 + x)(1 – x + x 2 )

And we know, a 3 + b 3 = (a + b)(a 2 + b 2 – ab)] (1 + x)(1 – x + x 2 ) can be written as, (1 + x)[(1 2 – (1)(x) + x 2 )] = (1) 3 + (x) 3 = 1 + x 3

(xi) (x 2 – 1)(x 4 + x 2 +1) can be written as,

(x 2 – 1)[(x 2 ) 2 – 1 2 + (x 2 )(1)] = (x 2 ) 3 – 1 3 = x 6 – 1 [using a 3 – b 3 = (a – b)(a 2 + b 2 + ab) ]

(xii) (x 3 + 1)(x 6 – x 3 + 1) can be written as,

(x 3 + 1)[(x 3 ) 2 – (x 3 )(1) + 1 2 ] = (x 3 ) 3 + 1 3 = x 9 + 1 [using a 3 + b 3 = (a + b)(a 2 + b 2 – ab) ]

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 )

(ii) (3/x – x/3)(x 2 /9 + 9/x 2 + 1)

(iii) (x/7 + y/3)(x 2 /49 + y 2 /9 – xy/21)

(iv) (x/4 – y/3)(x 2 /16 + xy/12 + y 2 /9) (v) (5/x + 5x)(25/x 2 – 25 + 25x 2 )

Solution:

(i) (9y 2 – 4x 2 )(81y 4 + 36x 2 y 2 + 16x 4 )

= (9y 2 – 4x 2 ) [(9y 2 ) 2 + 9y 2 x 4x 2 + (4x 2 ) 2 ] = (9y 2 ) 3 – (4x 2 ) 3 = 729 y 6 – 64 x 6 Put x = 3 and y = -1 = 729 – 46656 = – 45927

(ii) Put x = 3 and y = -1

(3/x – x/3)(x 2 /9 + 9/x 2 + 1) RD sharma class 9 maths chapter 4 ex 4.4 question 2 solution part 2

(iii) Put x = 3 and y = -1

(x/7 + y/3)(x 2 /49 + y 2 /9 – xy/21) RD sharma class 9 maths chapter 4 ex 4.4 question 2 solution part 3

(iv) Put x = 3 and y = -1

(x/4 – y/3)(x 2 /16 + xy/12 + y 2 /9) RD sharma class 9 maths chapter 4 ex 4.4 question 2 solution part 4

(v) Put x = 3 and y = -1

(5/x + 5x)(25/x 2 – 25 + 25x 2 ) RD sharma class 9 maths chapter 4 ex 4.4 question 2 solution part 5

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:

a + b = 10, ab = 16 Squaring, a + b = 10, both sides (a + b) 2 = (10) 2 a 2 + b 2 + 2ab = 100 a 2 + b 2 + 2 x 16 = 100 a 2 + b 2 + 32 = 100 a 2 + b 2 = 100 – 32 = 68 a 2 + b 2 = 68 Again, 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:

a + b = 8, ab = 6 Cubing, a + b = 8, both sides, we get (a + b) 3 = (8) 3 a 3 + b 3 + 3ab(a + b) = 512 a 3 + b 3 + 3 x 6 x 8 = 512 a 3 + b 3 + 144 = 512 a 3 + b 3 = 512 – 144 = 368 a 3 + b 3 = 368

RD Sharma Class 9 Solutions Maths Chapter 4 Algebraic Identities Exercise 4.5 Page No: 4.28

Question 1: Find the following products:

(i) (3x + 2y + 2z) (9x 2 + 4y 2 + 4z 2 – 6xy – 4yz – 6zx)

(ii) (4x – 3y + 2z) (16x 2 + 9y 2 + 4z 2 + 12xy + 6yz – 8zx)

(iii) (2a – 3b – 2c) (4a 2 + 9b 2 + 4c 2 + 6ab – 6bc + 4ca)

(iv) (3x -4y + 5z) (9x 2 + 16y 2 + 25z 2 + 12xy- 15zx + 20yz)

Solution:

(i) (3x + 2y + 2z) (9x 2 + 4y 2 + 4z 2 – 6xy – 4yz – 6zx)

= (3x + 2y + 2z) [(3x) 2 + (2y) 2 + (2z) 2 – 3x x 2y – 2y x 2z – 2z x 3x] = (3x) 3 + (2y) 3 + (2z) 3 – 3 x 3x x 2y x 2z = 27x 3 + 8y 3 + 8Z 3 – 36xyz

(ii) (4x – 3y + 2z) (16x 2 + 9y 2 + 4z 2 + 12xy + 6yz – 8zx)

= (4x -3y + 2z) [(4x) 2 + (-3y) 2 + (2z) 2 – 4x x (-3y) – (-3y) x (2z) – (2z x 4x)] = (4x) 3 + (-3y) 3 + (2z) 3 – 3 x 4x x (-3y) x (2z) = 64x 3 – 27y 3 + 8z 3 + 72xyz

(iii) (2a -3b- 2c) (4a 2 + 9b 2 + 4c 2 + 6ab – 6bc + 4ca)

= (2a -3b- 2c) [(2a) 2 + (-3b) 2 + (-2c) 2 – 2a x (-3b) – (-3b) x (-2c) – (-2c) x 2a] = (2a) 3 + (-3b) 3 + (-2c) 3 -3x 2a x (-3 b) (-2c) = 8a 3 – 21b 3 – 8c 3 – 36abc

(iv) (3x – 4y + 5z) (9x 2 + 16y 2 + 25z 2 + 12xy – 15zx + 20yz)

= [3x + (-4y) + 5z] [(3x) 2 + (-4y) 2 + (5z) 2 – 3x x (-4y) -(-4y) (5z) – 5z x 3x] = (3x) 3 + (-4y) 3 + (5z) 3 – 3 x 3x x (-4y) (5z) = 27x 3 – 64y 3 + 125z 3 + 180xyz

Question 2: If x + y + z = 8 and xy + yz+ zx = 20, find the value of x 3 + y 3 + z 3 – 3xyz.

Solution:

We know, x 3 + y 3 + z 3 – 3xyz = (x + y + z) (x 2 + y 2 + z 2 – xy – yz – zx) Squaring, x + y + z = 8 both sides, we get (x + y + z) 2 = (8) 2 x 2 + y 2 + z 2 + 2(xy + yz + zx) = 64 x 2 + y 2 + z 2 + 2 x 20 = 64 x 2 + y 2 + z 2 + 40 = 64 x 2 + y 2 + z 2 = 24 Now, x 3 + y 3 + z 3 – 3xyz = (x + y + z) [x 2 + y 2 + z 2 – (xy + yz + zx)] = 8(24 – 20) = 8 x 4 = 32 ⇒ x 3 + y 3 + z 3 – 3xyz = 32

Question 3: If a +b + c = 9 and ab + bc + ca = 26, find the value of a 3 + b 3 + c 3 – 3abc.

Solution:

a + b + c = 9, ab + bc + ca = 26 Squaring, a + b + c = 9 both sides, we get (a + b + c) 2 = (9) 2 a 2 + b 2 + c 2 + 2 (ab + bc + ca) = 81 a 2 + b 2 + c 2 + 2 x 26 = 81 a 2 + b 2 + c 2 + 52 = 81 a 2 + b 2 + c 2 = 29 Now, a 3 + b 3 + c 3 – 3abc = (a + b + c) [(a 2 + b 2 + c 2 – (ab + bc + ca)] = 9[29 – 26] = 9 x 3 = 27 ⇒ a 3 + b 3 + c 3 – 3abc = 27

RD Sharma Class 9 Solutions Maths Chapter 4 Algebraic Identities Exercise VSAQs Page No: 4.28

Question 1: If x + 1/x = 3, then find the value of x 2 + 1/x 2 .

Solution:

x + 1/x = 3 Squaring both sides, we have (x + 1/x) 2 = 3 2 x 2 + 1/x 2 + 2 = 9 x 2 + 1/x 2 = 9 – 2 = 7

Question 2: If x + 1/x = 3, then find the value of x^6 + 1/x^6.

Solution :

x + 1/x = 3 Squaring both sides, we have (x + 1/x) 2 = 3 2 x 2 + 1/x 2 + 2 = 9 x 2 + 1/x 2 = 9 – 2 = 7 x 2 + 1/x 2 = 7 …(1) Cubing equation (1) both sides, RD sharma class 9 maths chapter 4 ex vsaqs solutions

Question 3: If a + b = 7 and ab = 12, find the value of a 2 + b 2 .

Solution:

a + b = 7, ab = 12 Squaring, a + b = 7, both sides, (a + b) 2 = (7) 2 a 2 + b 2 + 2ab = 49 a 2 + b 2 + 2 x 12 = 49 a 2 + b 2 + 24 = 49 a 2 + b 2 = 25

Question 4: If a – b = 5 and ab = 12, find the value of a 2 + b 2 .

Solution:

a – b = 5, ab = 12 Squaring, a – b = 5, both sides, (a – b) 2 = (5) 2 a 2 + b 2 – 2ab = 25 a 2 + b 2 – 2 x 12 = 25 a 2 + b 2 – 24 = 25 a 2 + b 2 = 49
CBSE Class 9 Maths Syllabus CBSE Class 9 Science Syllabus
CBSE Class 9 Computer Application Syllabus CBSE Class 9 Social Science Syllabus

RD Sharma Solutions Class 9 Maths Chapter 4 FAQs

What topics are covered in RD Sharma Class 9 Maths Chapter 4 on Algebraic Identities?

RD Sharma Class 9 Maths Chapter 4 covers various topics related to algebraic identities, including polynomial identities, binomial theorem, and factorization.

How can RD Sharma Solutions help in understanding Algebraic Identities?

RD Sharma Solutions provide step-by-step explanations to problems, helping students understand the concepts better.

How can RD Sharma Solutions benefit students in exams?

By practicing with RD Sharma Solutions, students can improve their problem-solving skills and gain confidence in tackling exam questions. The solutions provide clarity on concepts and help students score well in exams.
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