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

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

## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 4 Algebraic Identities

### Access Answers to Maths RD Sharma Solutions for Class 9 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 }

^{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)

^{2}– b

^{2}]

(2x + y) (2x – y) = (2x )^{2 }– (y)^{2 }

= 4x^{2 } – y^{2 }

**(iii) **(a^{2}b – b^{2}a)^{2}

^{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)

^{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})

^{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 }

**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}

^{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}

^{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}

^{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) **

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

**Solution:**

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

**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) (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}

**(ix)** (a/bc + b/ac + c/ab)^{ 2}

**(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

### 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:**

^{3}= a

^{3}+ b

^{3}+ 3ab(a + b) and (a – b)

^{3}= a

^{3}– b

^{3}– 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: **

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)

**(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 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

**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

**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

**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

### 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})

**(v)** (3/x – 5/y)(9/x^{2} + 25/y^{2} + 15/xy)

^{3}– b

^{3}= (a – b)(a

^{2}+ b

^{2}+ ab) ]

**(vi)** (3 + 5/x)(9 – 15/x + 25/x^{2})

^{3}+ b

^{3}= (a + b)(a

^{2}+ b

^{2}– ab)]

**(vii)** (2/x + 3x)(4/x^{2 }+ 9x^{2 }– 6)

^{3}+ b

^{3}= (a + b)(a

^{2}+ b

^{2}– ab)]

**(viii)** (3/x – 2x^{2})(9/x^{2 }+ 4x^{4} – 6x)

^{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

^{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

^{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)

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

(x/7 + y/3)(x^{2}/49 + y^{2}/9 – xy/21)

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

(x/4 – y/3)(x^{2}/16 + xy/12 + y^{2}/9)

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

(5/x + 5x)(25/x^{2} – 25 + 25x^{2})

**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

### 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

### 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,

**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

### 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

These books are widely used by the students who wish to score high in board exams. For RD Sharma Class 9 Maths Solutions, you can visit the BYJU’S website, there you can get step by step answers to all the questions provided in RD Sharma textbook.

## Frequently Asked Questions on RD Sharma Solutions for Class 9 Maths Chapter 4

### Write the key benefits of RD Sharma Solutions for Class 9 Maths Chapter 4.

2. It also provides explanatory diagrams and tables for comparative study, which creates an interest in learning.

3. These solutions facilitate the students to build a good knowledge of basic as well as advanced mathematical concepts.

4. It also helps students in retaining and quickly retrieving the concepts.