RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities

RD Sharma Solutions Class 9 Maths is among the most popular study materials. Chapter 4 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.

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RD Sharma Solution class 9 Maths Chapter 4 Algebraic Identities 01
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Access Answers to Maths RD Sharma 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) (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 − 2bc − 2cb + 2ca) + (a2 + b2 + c2 + 2ab − 2bc – 2ab)

= 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]

= 4×2 + p2 + c2 + 4xp − 2pc − 4cx − 4×2 − 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 + 63 + 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 Exercises:

Get detailed solutions for all the questions listed under below exercises:

Exercise 4.1 Solutions

Exercise 4.2 Solutions

Exercise 4.3 Solutions

Exercise 4.4 Solutions

Exercise 4.5 Solutions

Exercise VSAQs Solutions

RD Sharma Solutions for Chapter 4 Algebraic Identities

In this chapter 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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