Class 8 Maths | Lesson 6: Factorisation of Algebraic Expressions | Practice Set 6.3 – Maharashtra State Board

 

📘 Lesson 6: Factorisation of Algebraic Expressions

Practice Set 6.3

📌 Formulae Used

a³ + b³ = (a + b)(a² − ab + b²)

a³ − b³ = (a − b)(a² + ab + b²)

✏️ Practice Set 6.3

1. Factorise

(1) y³ − 27

y³ − 27 = y³ − 3³

Using the formula:

a³ − b³ = (a − b)(a² + ab + b²)

Here,

a = y,   b = 3

y³ − 3³ = (y − 3)(y² + y(3) + 3²)

= (y − 3)(y² + 3y + 9)

∴ Answer = (y − 3)(y² + 3y + 9)

(2) x³ − 64y³

x³ − 64y³ = x³ − (4y)³

Here,

a = x,   b = 4y

Using the formula:

= (x − 4y){x² + x(4y) + (4y)²}

= (x − 4y)(x² + 4xy + 16y²)

∴ Answer = (x − 4y)(x² + 4xy + 16y²)

(3) 27m³ − 216n³

Take 27 common:

27m³ − 216n³ = 27(m³ − 8n³)

= 27{m³ − (2n)³}

Here,

a = m,   b = 2n

Using the formula:

= 27(m − 2n){m² + m(2n) + (2n)²}

= 27(m − 2n)(m² + 2mn + 4n²)

∴ Answer = 27(m − 2n)(m² + 2mn + 4n²)

(4) 125y³ − 1

125y³ − 1 = (5y)³ − 1³

Here,

a = 5y,   b = 1

Using the formula:

= (5y − 1){(5y)² + (5y)(1) + 1²}

= (5y − 1)(25y² + 5y + 1)

∴ Answer = (5y − 1)(25y² + 5y + 1)

(5) 8p³ − 27/p³

8p³ − 27/p³ = (2p)³ − (3/p)³

Here,

a = 2p,   b = 3/p

Using the formula:

= (2p − 3/p) {(2p)² + (2p)(3/p) + (3/p)²}

= (2p − 3/p) (4p² + 6 + 9/p²)

∴ Answer = (2p − 3/p)(4p² + 6 + 9/p²)

(6) 343a³ − 512b³

343a³ − 512b³ = (7a)³ − (8b)³

Here,

a = 7a,   b = 8b

Using the formula:

= (7a − 8b){(7a)² + (7a)(8b) + (8b)²}

= (7a − 8b)(49a² + 56ab + 64b²)

∴ Answer = (7a − 8b)(49a² + 56ab + 64b²)

(7) 64x³ − 729y³

64x³ − 729y³ = (4x)³ − (9y)³

Here,

a = 4x,   b = 9y

Using the formula:

= (4x − 9y){(4x)² + (4x)(9y) + (9y)²}

= (4x − 9y)(16x² + 36xy + 81y²)

∴ Answer = (4x − 9y)(16x² + 36xy + 81y²)

(8) 16a³ − 128/b³

Take 16 common:

16a³ − 128/b³ = 16(a³ − 8/b³)

= 16{a³ − (2/b)³}

Here,

a = a,   b = 2/b

Using the formula:

= 16(a − 2/b) {a² + a(2/b) + (2/b)²}

= 16(a − 2/b)(a² + 2a/b + 4/b²)

∴ Answer = 16(a − 2/b)(a² + 2a/b + 4/b²)

2. Simplify

(1) (x + y)³ − (x − y)³

Using the identity:

(a + b)³ − (a − b)³ = 6a²b + 2b³

Here,

a = x,   b = y

= 6x²y + 2y³

= 2y(3x² + y²)

∴ Answer = 2y(3x² + y²)

(2) (3a + 5b)³ − (3a − 5b)³

Using:

(A + B)³ − (A − B)³ = 6A²B + 2B³

Here,

A = 3a,   B = 5b

= 6(3a)²(5b) + 2(5b)³

= 6(9a²)(5b) + 2(125b³)

= 270a²b + 250b³

= 10b(27a² + 25b²)

∴ Answer = 10b(27a² + 25b²)

(3) (a + b)³ − a³ − b³

Expand:

(a + b)³ = a³ + 3a²b + 3ab² + b³

Therefore,

(a + b)³ − a³ − b³

= a³ + 3a²b + 3ab² + b³ − a³ − b³

= 3a²b + 3ab²

= 3ab(a + b)

∴ Answer = 3ab(a + b)

(4) p³ − (p + 1)³

Expand:

(p + 1)³ = p³ + 3p² + 3p + 1

Therefore,

p³ − (p + 1)³

= p³ − (p³ + 3p² + 3p + 1)

= p³ − p³ − 3p² − 3p − 1

= −3p² − 3p − 1

∴ Answer = −3p² − 3p − 1

(5) (3xy − 2ab)³ − (3xy + 2ab)³

Let

A = 3xy,   B = 2ab

Using:

(A − B)³ − (A + B)³ = −6A²B − 2B³

= −6(3xy)²(2ab) − 2(2ab)³

= −6(9x²y²)(2ab) − 2(8a³b³)

= −108abx²y² − 16a³b³

= −4ab(27x²y² + 4a²b²)

∴ Answer = −4ab(27x²y² + 4a²b²)

✅ Final Answers

Question	Answer
1 (1)	(y − 3)(y² + 3y + 9)
1 (2)	(x − 4y)(x² + 4xy + 16y²)
1 (3)	27(m − 2n)(m² + 2mn + 4n²)
1 (4)	(5y − 1)(25y² + 5y + 1)
1 (5)	(2p − 3/p)(4p² + 6 + 9/p²)
1 (6)	(7a − 8b)(49a² + 56ab + 64b²)
1 (7)	(4x − 9y)(16x² + 36xy + 81y²)
1 (8)	16(a − 2/b)(a² + 2a/b + 4/b²)
2 (1)	2y(3x² + y²)
2 (2)	10b(27a² + 25b²)
2 (3)	3ab(a + b)
2 (4)	−3p² − 3p − 1
2 (5)	−4ab(27x²y² + 4a²b²)