RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities

RD Sharma Solutions for Class 9 Maths Chapter 4 – Free PDF Download

RD Sharma Solutions Class 9 Chapter 4 is among the most popular study materials. Algebraic identities is a detailed study material prepared by subject experts at BYJU’S. It provides answers to the questions given in the textbook. Everything is presented in a way that students can easily understand. The algebraic equations which are true for all values of variables in them are called algebraic identities. RD Sharma Class 9 Solutions are helpful to practice on factorization of polynomials. It will also help students score well in the examination.

Students who aim to secure good marks in their board marks can refer to RD Sharma Solutions. These solutions are formulated by the BYJU’S expert team in Maths to help students solve the problems. Students who find it difficult in solving exercise wise problems of RD Sharma textbook can access PDF of solutions.

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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) (a2b – b2a)2

(iv) (a – 0.1) (a + 0.1)

(v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2)

Solution:

(i) (2x – 1/x)2

[Use identity: (a – b)2 = a2 + b2 – 2ab ]

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

= 4x2 + 1/x2 – 4

(ii) (2x + y) (2x – y)

[Use identity: (a – b)(a + b) = a2 – b2 ]

(2x + y) (2x – y) = (2x )2 – (y)2

= 4x2 – y2

(iii) (a2b – b2a)2

[Use identity: (a – b)2 = a2 + b2 – 2ab ]

 

(a2b – b2a)2 = (a2b) 2 + (b2a)2 – 2 (a2b)( b2a)

= a4b 2 + b4a2 – 2 a3b3

(iv) (a – 0.1) (a + 0.1)

[Use identity: (a – b)(a + b) = a2 – b2 ]

(a – 0.1) (a + 0.1) = (a)2 – (0.1)2

= (a)2 – 0.01

(v) (1.5 x2 – 0.3y2) (1.5 x2 + 0.3y2)

[Use identity: (a – b)(a + b) = a2 – b2 ]

(1.5 x2 – 0.3y2) (1.5x2 + 0.3y2) = (1.5 x2 ) 2 – (0.3y2)2

= 2.25 x4 – 0.09y4

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 a2+ b2+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 a2+ b2-2ab = (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 a2+ b2+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 x2 +1/x2.

Solution:

RD sharma class 9 maths chapter 4 ex 4.1 question 4

Question 5: If x – 1/x = -1, find the value of x2 +1/x2.

Solution:

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


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) (a2 +b2 +c2) 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 = x2 + y2 + z2 + 2xy + 2yz + 2xz

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

= a2 + (2b) 2 + c2 + 2a(2b) + 2ac + 2(2b)c

= a2 + 4b2 + c2 + 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)

= 4a2 + 9b2 + c2 − 12ab + 6bc − 4ca

(iii) (−3x+y+z)2

= [(−3x) 2 + y2 + z2 + 2(−3x)y + 2yz + 2(−3x)z

= 9x2 + y2 + z2 − 6xy + 2yz − 6xz

(iv) (m+2n−5p)2

= m2 + (2n) 2 + (−5p) 2 + 2m × 2n + (2×2n×−5p) + 2m × −5p

= m2 + 4n2 + 25p2 + 4mn − 20np − 10pm

(v) (2+x−2y)2

= 22 + x2 + (−2y) 2 + 2(2)(x) + 2(x)(−2y) + 2(2)(−2y)

= 4 + x2 + 4y2 + 4 x − 4xy − 8y

(vi) (a2 +b2 +c2) 2

= (a2) 2 + (b2) 2 + (c2 ) 2 + 2a2 b2 + 2b2c2 + 2a2c2

= a4 + b4 + c4 + 2a2 b2 + 2b2 c2 + 2c2 a2

(vii) (ab+bc+ca) 2

= (ab)2 + (bc) 2 + (ca) 2 + 2(ab)(bc) + 2(bc)(ca) + 2(ab)(ca)

= a2 b2 + b2c2 + c2 a2 + 2(ac)b2 + 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

= x2 + (2y) 2 + (4z) 2 + (2x)(2y) + 2(2y)(4z) + 2x(4z)

= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(xi) (2x−y+z) 2

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

= 4x2 + y2 + z2 − 4xy−2yz+4xz

(xii) (−2x+3y+2z) 2

= (−2x) 2 + (3y) 2 + ( 2z) 2 + 2(−2x)(3y)+2(3y)(2z)+2(−2x)(2z)

= 4x2 + 9y2 + 4z2 −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) (x2 + y2 − z2) 2 − (x2 − y2 + z2) 2

Solution:

(i) (a + b + c)2 + (a − b + c) 2

= (a2 + b2 + c2 + 2ab+2bc+2ca) + (a2 + (−b) 2 + c2 −2ab−2bc+2ca)

= 2a2 + 2 b2 + 2c2 + 4ca

(ii) (a + b + c)2 − (a − b + c) 2

= (a2 + b2 + c2 + 2ab+2bc+2ca) − (a2 + (−b) 2 + c2 −2ab−2bc+2ca)

= a2 + b2 + c2 + 2ab + 2bc + 2ca − a2 − b2 − c2 + 2ab + 2bc − 2ca

= 4ab + 4bc

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

= a2 + b2 + c2 + 2ab + 2bc + 2ca + (a2 + b2 + (c) 2 − 2ab − 2cb + 2ca) + (a2 + b2 + c2 + 2ab − 2bc – 2ca)

= 3 a2 + 3b2 + 3c2 + 2ab − 2bc + 2ca

(iv) (2x + p − c)2 − (2x − p + c) 2

= [4x2 + p2 + c2 + 4xp − 2pc − 4xc] − [4x2 + p2 + c2 − 4xp− 2pc + 4xc]

= 4x2 + p2 + c2 + 4xp − 2pc − 4cx − 4x2 − p2 − c2 + 4xp + 2pc− 4cx

= 8xp − 8xc

= 8(xp − xc)

(v) (x2 + y2 − z2) 2 − (x2 − y2 + z2) 2

= (x2 + y2 + (−z) 2) 2 − (x2 − y2 + z2) 2

= [x4 + y4 + z4 + 2x2y2 – 2y2z 2 – 2x2z2 − [x4 + y4 + z4 − 2x2y2 − 2y2z2 + 2x2z2]

= 4x2y2 – 4z2x2

Question 3: If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + bc + ca.

Solution:

a + b + c = 0 and a2 + b2 + c2 = 16 (given)

Choose a + b + c = 0

Squaring both sides,

(a + b + c)2 = 0

a2 + b2 + c2 + 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


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/x2)

(iii) (2x + 3/x)

(iv) (4 – 1/3x)

Solution:

[Using identities: (a + b)3 = a3 + b3 + 3ab(a + b) and (a – b)3 = a3 – b3 – 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:

a3 + b3 = (a + b)(a2 + b2 – ab)

a3 – b3 = (a – b)(a2 + b2 + ab)

(a + b)(a-b) = a2 – b2

(a + b)2 = a2 + b2 + 2ab and

(a – b)2 = a2 + b2 – 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 a3 + b3.

Solution:

a + b = 10, ab = 21 (given)

Choose a + b = 10

Cubing both sides,

(a + b)3 = (10)3

a3 + b3 + 3ab(a + b) = 1000

a3 + b3 + 3 x 21 x 10 = 1000 (using given values)

a3 + b3 + 630 = 1000

a3 + b3 = 1000 – 630 = 370

or a3 + b3 = 370

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

Solution:

a – b = 4, ab= 21 (given)

Choose a – b = 4

Cubing both sides,

(a – b)3 = (4)3

a3 – b3 – 3ab (a – b) = 64

a3 – b3 – 3 × 21 x 4 = 64 (using given values)

a3 – b3 – 252 = 64

a3 – b3 = 64 + 252

= 316

Or a3 – b3 = 316

Question 5: If x + 1/x = 5, find the value of x3 + 1/x3 .

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 x3 – 1/x3 .

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 x3 – 1/x3 .

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 (x2 + 1/x2) = 51, find the value of x3 – 1/x3.

Solution:

We know that: (x – y)2 = x2 + y2 – 2xy

Replace y with 1/x, we get

(x – 1/x)2 = x2 + 1/x2 – 2

Since (x2 + 1/x2) = 51 (given)

(x – 1/x)2 = 51 – 2 = 49

or (x – 1/x) = ±7

Now, Find x3 – 1/x3

We know that, x3 – y3 = (x – y)(x2 + y2 + xy)

Replace y with 1/x, we get

x3 – 1/x3 = (x – 1/x)(x2 + 1/x2 + 1)

Use (x – 1/x) = 7 and (x2 + 1/x2) = 51

x3 – 1/x3 = 7 x 52 = 364

x3 – 1/x3 = 364

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

Solution:

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

Replace y with 1/x, we get

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

Since (x2 + 1/x2) = 98 (given)

(x + 1/x)2 = 98 + 2 = 100

or (x + 1/x) = ±10

Now, Find x3 + 1/x3

We know that, x3 + y3 = (x + y)(x2 + y2 – xy)

Replace y with 1/x, we get

x3 + 1/x3 = (x + 1/x)(x2 + 1/x2 – 1)

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

x3 + 1/x3 = 10 x 97 = 970

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

Cubing 2x + 3y = 13 both sides, we get

(2x + 3y)3 = (13)3

(2x)3 + (3y) 3 + 3( 2x )(3y) (2x + 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 11: If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3.

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

27x3 – 8y3 – 18xy(3x -2y) =1331

27x3 – 8y3 – 18 x 12 x 11 = 1331

27x3 – 8y3 – 2376 = 1331

27x3 – 8y3 = 1331 + 2376 = 3707

27x3 – 8y3 = 3707


Exercise 4.4 Page No: 4.23

Question 1: Find the following products:

(i) (3x + 2y)(9x2 – 6xy + 4y2)

(ii) (4x – 5y)(16x2 + 20xy + 25y2)

(iii) (7p4 + q)(49p8 – 7p4q + q2)

(iv) (x/2 + 2y)(x2/4 – xy + 4y2)

(v) (3/x – 5/y)(9/x2 + 25/y2 + 15/xy)

(vi) (3 + 5/x)(9 – 15/x + 25/x2)

(vii) (2/x + 3x)(4/x2 + 9x2 – 6)

(viii) (3/x – 2x2)(9/x2 + 4x4 – 6x)

(ix) (1 – x)(1 + x + x2)

(x) (1 + x)(1 – x + x2)

(xi) (x2 – 1)(x4 + x2 +1)

(xii) (x3 + 1)(x6 – x3 + 1)

Solution:

(i) (3x + 2y)(9x2 – 6xy + 4y2)

= (3x + 2y)[(3x)2 – (3x)(2y) + (2y)2)]

We know, a3 + b3 = (a + b)(a2 + b2 – ab)

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

= 27x3 + 8y3

(ii) (4x – 5y)(16x2 + 20xy + 25y2)

= (4x – 5y)[(4x)2 + (4x)(5y) + (5y)2)]

We know, a3 – b3 = (a – b)(a2 + b2 + ab)

= (4x)3 – (5y) 3

= 64x3 – 125y3

(iii) (7p4 + q)(49p8 – 7p4q + q2)

= (7p4 + q)[(7p4)2 – (7p4)(q) + (q)2)]

We know, a3 + b3 = (a + b)(a2 + b2 – ab)

= (7p4)3 + (q) 3

= 343 p12 + q3

(iv) (x/2 + 2y)(x2/4 – xy + 4y2)

We know, a3 – b3 = (a – b)(a2 + b2 + ab)

(x/2 + 2y)(x2/4 – xy + 4y2)

RD sharma class 9 maths chapter 4 ex 4.4 question 2 Solution

(v) (3/x – 5/y)(9/x2 + 25/y2 + 15/xy)

RD sharma class 9 maths chapter 4 ex 4.4

[Using a3 – b3 = (a – b)(a2 + b2 + ab) ]

(vi) (3 + 5/x)(9 – 15/x + 25/x2)

RD sharma class 9 maths chapter 4 ex 4.4 solution

[Using: a3 + b3 = (a + b)(a2 + b2 – ab)]

(vii) (2/x + 3x)(4/x2 + 9x2 – 6)

RD sharma class 9 maths chapter 4 ex 4.4 question 1

[Using: a3 + b3 = (a + b)(a2 + b2 – ab)]

(viii) (3/x – 2x2)(9/x2 + 4x4 – 6x)

RD sharma class 9 maths chapter 4 ex 4.4 question 1 solution

[Using : a3 – b3 = (a – b)(a2 + b2 + ab)]

(ix) (1 – x)(1 + x + x2)

And we know, a3 – b3 = (a – b)(a2 + b2 + ab)

(1 – x)(1 + x + x2) can be written as

(1 – x)[(12 + (1)(x)+ x2)]

= (1)3 – (x)3

= 1 – x3

(x) (1 + x)(1 – x + x2)

And we know, a3 + b3 = (a + b)(a2 + b2 – ab)]

(1 + x)(1 – x + x2) can be written as,

(1 + x)[(12 – (1)(x) + x2)]

= (1)3 + (x) 3

= 1 + x3

(xi) (x2 – 1)(x4 + x2 +1) can be written as,

(x2 – 1)[(x2)2 – 12 + (x2)(1)]

= (x2)3 – 13

= x6 – 1

[using a3 – b3 = (a – b)(a2 + b2 + ab) ]

(xii) (x3 + 1)(x6 – x3 + 1) can be written as,

(x3 + 1)[(x3)2 – (x3)(1) + 12]

= (x3) 3 + 13

= x9 + 1

[using a3 + b3 = (a + b)(a2 + b2 – ab) ]

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

(i) (9y2 – 4x2)(81y4 + 36x2y2 + 16x4)

(ii) (3/x – x/3)(x2 /9 + 9/x2 + 1)

(iii) (x/7 + y/3)(x2/49 + y2/9 – xy/21)

(iv) (x/4 – y/3)(x2/16 + xy/12 + y2/9)
(v) (5/x + 5x)(25/x2 – 25 + 25x2)

Solution:

(i) (9y2 – 4x2)(81y4 + 36x2y2 + 16x4)

= (9y2 – 4x2) [(9y2 ) 2 + 9y2 x 4x2 + (4x2) 2 ]

= (9y2 ) 3 – (4x2)3

= 729 y6 – 64 x6

Put x = 3 and y = -1

= 729 – 46656

= – 45927

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

(3/x – x/3)(x2 /9 + 9/x2 + 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)(x2/49 + y2/9 – 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)(x2/16 + xy/12 + y2/9)

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/x2 – 25 + 25x2)

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 a2 – ab + b2 and a2 + ab + b2.

Solution:

a + b = 10, ab = 16

Squaring, a + b = 10, both sides

(a + b)2 = (10)2

a2 + b2 + 2ab = 100

a2 + b2 + 2 x 16 = 100

a2 + b2 + 32 = 100

a2 + b2 = 100 – 32 = 68

a2 + b2 = 68

Again, a2 – ab + b2 = a2 + b2 – ab = 68 – 16 = 52 and

a2 + ab + b2 = a2 + b2 + ab = 68 + 16 = 84

Question 4: If a + b = 8 and ab = 6, find the value of a3 + b3.

Solution:

a + b = 8, ab = 6

Cubing, a + b = 8, both sides, we get

(a + b)3 = (8)3

a3 + b3 + 3ab(a + b) = 512

a3 + b3 + 3 x 6 x 8 = 512

a3 + b3 + 144 = 512

a3 + b3 = 512 – 144 = 368

a3 + b3 = 368


Exercise 4.5 Page No: 4.28

Question 1: Find the following products:

(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)

(ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx)

(iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)

(iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz)

Solution:

(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 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

= 27x3 + 8y3 + 8Z3 – 36xyz

(ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 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)

= 64x3 – 27y3 + 8z3 + 72xyz

(iii) (2a -3b- 2c) (4a2 + 9b2 + 4c2 + 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)

= 8a3 – 21b3 – 8c3 – 36abc

(iv) (3x – 4y + 5z) (9x2 + 16y2 + 25z2 + 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)

= 27x3 – 64y3 + 125z3 + 180xyz

Question 2: If x + y + z = 8 and xy + yz+ zx = 20, find the value of x3 + y3 + z3 – 3xyz.

Solution:

We know, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

Squaring, x + y + z = 8 both sides, we get

(x + y + z)2 = (8) 2

x2 + y2 + z2 + 2(xy + yz + zx) = 64

x2 + y2 + z2 + 2 x 20 = 64

x2 + y2 + z2 + 40 = 64

x2 + y2 + z2 = 24

Now,

x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)]

= 8(24 – 20)

= 8 x 4

= 32

⇒ x3 + y3 + z3 – 3xyz = 32

Question 3: If a +b + c = 9 and ab + bc + ca = 26, find the value of a3 + b3 + c3 – 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

a2 + b2 + c2 + 2 (ab + bc + ca) = 81

a2 + b2 + c2 + 2 x 26 = 81

a2 + b2 + c2 + 52 = 81

a2 + b2 + c2 = 29

Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)]

= 9[29 – 26]

= 9 x 3

= 27

⇒ a3 + b3 + c3 – 3abc = 27


Exercise VSAQs Page No: 4.28

Question 1: If x + 1/x = 3, then find the value of x2 + 1/x2.

Solution:

x + 1/x = 3

Squaring both sides, we have

(x + 1/x)2 = 32

x2 + 1/x2 + 2 = 9

x2 + 1/x2 = 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 = 32

x2 + 1/x2 + 2 = 9

x2 + 1/x2 = 9 – 2 = 7

x2 + 1/x2 = 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 a2 + b2.

Solution:

a + b = 7, ab = 12

Squaring, a + b = 7, both sides,

(a + b) 2 = (7) 2

a2 + b2 + 2ab = 49

a2 + b2 + 2 x 12 = 49

a2 + b2 + 24 = 49

a2 + b2 = 25

Question 4: If a – b = 5 and ab = 12, find the value of a2 + b2.

Solution:

a – b = 5, ab = 12

Squaring, a – b = 5, both sides,

(a – b)2 = (5)2

a2 + b2 – 2ab = 25

a2 + b2 – 2 x 12 = 25

a2 + b2 – 24 = 25

a2 + b2 = 49

RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities

In the 4th chapter of Class 9 RD Sharma Solutions students will study important identities as listed below:

  • Algebraic Identities Introduction
  • Identity for the square of a trinomial
  • Sum and difference of cubes Identity

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