RD Sharma Solutions for Class 9 Maths Chapter 4–Algebraic Identities
RD Sharma Solutions for Class 9 Maths Chapter 4–Algebraic Identities

Class 9: Maths Chapter 4 solutions. Complete Class 9 Maths Chapter 4 Notes.

RD Sharma Solutions for Class 9 Maths Chapter 4–Algebraic Identities

RD Sharma 9th Maths Chapter 4, Class 9 Maths Chapter 4 solutions

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)

(ii)

(iii)

(iv)

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)

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)

Question 4: If x + 1/x = 11, find the value of x2 +1/x2.

Solution:

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

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

(ix) (a/bc + b/ac + c/ab) 2

(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)

(ii)

(iii)

(iv)

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)

(ii) (x/2 + y/3) 3 – (x/2 – y/3) 3

Here a = (x/2 + y/3) and b = (x/2 – y/3)

(iii) (x + 2/x) 3 + (x – 2/x) 3

Here a = (x + 2/x) and b = (x – 2/x)

(iv) (2x – 5y) 3 – (2x + 5y) 3

Here a = (2x – 5y) and b = 2x + 5y

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

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

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

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)

(v) (3/x – 5/y)(9/x2 + 25/y2 + 15/xy)

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

(vi) (3 + 5/x)(9 – 15/x + 25/x2)

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

(vii) (2/x + 3x)(4/x2 + 9x2 – 6)

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

(viii) (3/x – 2x2)(9/x2 + 4x4 – 6x)

[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)

(iii) Put x = 3 and y = -1

(x/7 + y/3)(x2/49 + y2/9 – xy/21)

(iv) Put x = 3 and y = -1

(x/4 – y/3)(x2/16 + xy/12 + y2/9)

(v) Put x = 3 and y = -1

(5/x + 5x)(25/x2 – 25 + 25x2)

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,

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: Download PDF

RD Sharma Solutions for Class 9 Maths Chapter 4–Algebraic Identities

Download PDF: RD Sharma Solutions for Class 9 Maths Chapter 4–Algebraic Identities PDF

Chapterwise RD Sharma Solutions for Class 9 Maths :

About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.

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