Factorisation

   4x² + 16

= 4(x² + 4).

Hence, Option 1 is the correct option.

   7(x + 3y)² - 2x - 6y

= 7(x + 3y)² - 2(x + 3y)

= (x + 3y)[7(x + 3y) - 2]

= (x + 3y)(7x + 21y - 2)

Hence, Option 3 is the correct option.

   8(x - y) + 5(y - x)

= 8(x - y) - 5(x - y)

= (x - y)[8 - 5]

= 3(x - y).

Hence, Option 3 is the correct option.

$$\begin{aligned}\phantom{\Rightarrow} 4x^2 + \frac{1}{4x^2} - 2 - 6x + \frac{3}{2x} \\[1em] = \Big(4x^2 + \frac{1}{4x^2} - 2\Big) - 6x + \frac{3}{2x} \\[1em] = \Big(2x - \frac{1}{2x}\Big)^2 - 3\Big(2x - \frac{1}{2x}\Big) \\[1em] = \Big(2x - \frac{1}{2x}\Big)\Big(2x - \frac{1}{2x} - 3\Big)\end{aligned}$$

Hence, Option 4 is the correct option.

  3ab - 6b + 4a² - 8a

= 3b(a - 2) + 4a(a - 2)

= (a - 2)(3b + 4a).

Hence, Option 2 is the correct option.

Given,

xy(3x² - 2y²) - yz(2y² - 3x²) + zx(15x² - 10y²)

= xy(3x² - 2y²) - yz[-(3x² - 2y²)] + 5zx(3x² - 2y²)

= xy(3x² - 2y²) + yz(3x² - 2y²) + 5zx(3x² - 2y²)

= (3x² - 2y²)(xy + yz + 5zx).

Hence, xy(3x² - 2y²) - yz(2y² - 3x²) + zx(15x² - 10y²) = (3x² - 2y²)(xy + yz + 5zx).

Given,

   2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)

= 2x(a - b) + 3y × 5(a - b) + 4z × -2(a - b)

= 2x(a - b) + 15y(a - b) - 8z(a - b)

= (a - b)(2x + 15y - 8z).

Hence, 2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a) = (a - b)(2x + 15y - 8z).

Given,

   16(a + b)² - 4a - 4b

= 16(a + b)² - 4(a + b)

= 4(a + b)[4(a + b) - 1]

= 4(a + b)[4a + 4b - 1]

= 4(a + b)(4a + 4b - 1).

Hence, 16(a + b)² - 4a - 4b = 4(a + b)(4a + 4b - 1).

Given,

   a⁴ - 2a³ - 4a + 8

= a⁴ - 4a - 2a³ + 8

= a(a³ - 4) - 2(a³ - 4)

= (a - 2)(a³ - 4).

Hence, a⁴ - 2a³ - 4a + 8 = (a - 2)(a³ - 4).

Given,

   ab(x² + 1) + x(a² + b²)

= abx² + ab + xa² + xb²

= abx² + xa² + xb² + ab

= ax(bx + a) + b(bx + a)

= (ax + b)(bx + a).

Hence, ab(x² + 1) + x(a² + b²) = (ax + b)(bx + a).

Given,

   (ax + by)² + (bx - ay)²

= (ax)² + (by)² + 2 × ax × by + (bx)² + (ay)² - 2 × bx × ay

= a²x² + b²y² + 2abxy + b²x² + a²y² - 2abxy

= a²x² + b²x² + a²y² + b²y²

= x²(a² + b²) + y²(a² + b²)

= (x² + y²)(a² + b²).

Hence, (ax + by)² + (bx - ay)² = (x² + y²)(a² + b²).

Given,

   a²x² + (ax² + 1)x + a

= a²x² + ax³ + x + a

= a²x² + ax³ + a + x

= ax²(a + x) + 1(a + x)

= (a + x)(ax² + 1).

Hence, a²x² + (ax² + 1)x + a = (a + x)(ax² + 1).

Given,

   y² - (a + b)y + ab

= y² - ay - by + ab

= y(y - a) - b(y - a)

= (y - a)(y - b).

Hence, y² - (a + b)y + ab = (y - a)(y - b).

Given,

   x² + 2(5x + 12)

= x² + 10x + 24

= x² + 6x + 4x + 24

= x(x + 6) + 4(x + 6)

= (x + 4)(x + 6).

Hence, Option 3 is the correct option.

Given,

   x(2x - 5) + 3

= 2x² - 5x + 3

= 2x² - 3x - 2x + 3

= x(2x - 3) - 1(2x - 3)

= (x - 1)(2x - 3).

Hence, Option 4 is the correct option.

Given,

   6x² - 2 - 4x

= 6x² - 4x - 2

= 6x² - 6x + 2x - 2

= 6x(x - 1) + 2(x - 1)

= (x - 1)(6x + 2)

= 2(x - 1)(3x + 1).

Hence, Option 2 is the correct option.

Given,

   x(x - 1) - 20

= x² - x - 20

= x² - 5x + 4x - 20

= x(x - 5) + 4(x - 5)

= (x - 5)(x + 4).

Hence, Option 1 is the correct option.

Given,

   5 - x(6 - x)

= 5 - 6x + x²

= x² - 6x + 5

= x² - 5x - x + 5

= x(x - 5) - 1(x - 5)

= (x - 1)(x - 5)

= [-(1 - x).-(5 - x)]

= (1 - x)(5 - x).

Hence, Option 3 is the correct option.

Given,

   1 - 2a - 3a²

= -(3a² + 2a - 1)

= -[3a² + 3a - a - 1]

= -[3a(a + 1) - 1(a + 1)]

= -(3a - 1)(a + 1)

= (1 - 3a)(a + 1).

Hence, 1 - 2a - 3a² = (1 - 3a)(a + 1).

Given,

   24a³ + 37a² - 5a

= 24a³ + 40a² - 3a² - 5a

= 8a²(3a + 5) - a(3a + 5)

= (8a² - a)(3a + 5)

= a(8a - 1)(3a + 5).

Hence, 24a³ + 37a² - 5a = a(8a - 1)(3a + 5).

Given,

   3 - a(4 + 7a)

= 3 - 4a - 7a²

= -7a² - 4a + 3

= -(7a² + 4a - 3)

= -[7a² + 7a - 3a - 3]

= -[7a(a + 1) - 3(a + 1)]

= -[(a + 1)(7a - 3)]

= -[-(a + 1)(3 - 7a)]

= (a + 1)(3 - 7a).

Hence, 3 - a(4 + 7a) = (a + 1)(3 - 7a).

Given,

   (2a + b)² - 6a - 3b - 4

= (2a + b)² - 3(2a + b) - 4

Substituting (2a + b) = x, we get :

= x² - 3x - 4

= x² - 4x + x - 4

= x(x - 4) + 1(x - 4)

= (x - 4)(x + 1)

= (2a + b - 4)(2a + b + 1).

Hence, (2a + b)² - 6a - 3b - 4 = (2a + b - 4)(2a + b + 1).

Given,

   1 - 2a - 2b - 3(a + b)²

= 1 - 2(a + b) - 3(a + b)²

Substituting (a + b) = x, we get :

= 1 - 2x - 3x²

= -3x² - 2x + 1

= -[3x² + 2x - 1]

= -[3x² + 3x - x - 1]

= -[3x(x + 1) - 1(x + 1)]

= -[(x + 1)(3x - 1)]

= -[-(x + 1)(1 - 3x)]

= (x + 1)(1 - 3x)

= (a + b + 1)[1 - 3(a + b)]

= (a + b + 1)(1 - 3a - 3b).

Hence, 1 - 2a - 2b - 3(a + b)² = (a + b + 1)(1 - 3a - 3b).

Given,

   3a² - 1 - 2a

= 3a² - 2a - 1

= 3a² - 3a + a - 1

= 3a(a - 1) + 1(a - 1)

= (a - 1)(3a + 1).

Hence, 3a² - 1 - 2a = (a - 1)(3a + 1).

Given,

   x² + 3x + 2 + ax + 2a

= x² + 3x + 2 + a(x + 2)

= x² + 2x + x + 2 + a(x + 2)

= x(x + 2) + 1(x + 2) + a(x + 2)

= (x + 2)(x + 1 + a).

Hence, x² + 3x + 2 + ax + 2a = (x + 2)(x + 1 + a).

Given,

   (3x - 2y)² + 3(3x - 2y) - 10

Substituting (3x - 2y) = a, we get :

⇒ a² + 3a - 10

= a² + 5a - 2a - 10

= a(a + 5) - 2(a + 5)

= (a + 5)(a - 2)

= (3x - 2y + 5)(3x - 2y - 2).

Hence, (3x - 2y)² + 3(3x - 2y) - 10 = (3x - 2y + 5)(3x - 2y - 2).

Given,

   5 - (3a² - 2a)(6 - 3a² + 2a)

= 5 - [-(3a² - 2a)(3a² - 2a - 6)]

= 5 + (3a² - 2a)(3a² - 2a - 6)

Substituting 3a² - 2a = x, we get :

= 5 + x(x - 6)

= 5 + x² - 6x

= x² - 6x + 5

= x² - 5x - x + 5

= x(x - 5) - 1(x - 5)

= (x - 5)(x - 1)

= (3a² - 2a - 5)(3a² - 2a - 1)

= [3a² - 5a + 3a - 5] [3a² - 3a + a - 1]

= [a(3a - 5) + 1(3a - 5)] [3a(a - 1) + 1(a - 1)]

= (3a - 5)(a + 1)(a - 1)(3a + 1).

Hence, 5 - (3a² - 2a)(6 - 3a² + 2a) = (3a - 5)(a + 1)(a - 1)(3a + 1).

Given,

   x⁴ - 1

= (x²)² - 1²

= (x² - 1)(x² + 1)

= (x - 1)(x + 1)(x² + 1).

Hence, Option 2 is the correct option.

Given,

   2a² - 18

= 2(a² - 9)

= 2[(a)² - (3)²]

= 2(a + 3)(a - 3).

Hence, Option 1 is the correct option.

Given,

   27x³ - 48x

= 3x(9x² - 16)

= 3x[(3x)² - (4)²]

= 3x(3x - 4)(3x + 4).

Hence, Option 4 is the correct option.

Given,

   x² - (x - 4y)²

= [x + (x - 4y)][x - (x - 4y)]

= (2x - 4y)[x - x + 4y]

= 4y(2x - 4y)

= 8y(x - 2y).

Hence, Option 1 is the correct option.

Given,

$$\begin{aligned}\phantom{=} 9x^2 + \frac{1}{9x^2} + 1 \\[1em] = 9x^2 + \frac{1}{9x^2} + 2 - 1 \\[1em] = (3x)^2 + \Big(\frac{1}{3x}\Big)^2 + 2 - 1 \\[1em] = \Big(3x + \frac{1}{3x}\Big)^2 - 1^2 \\[1em] = \Big(3x + \frac{1}{3x} + 1\Big)\Big(3x + \frac{1}{3x} - 1\Big).\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

   a² - (2a + 3b)²

= (a + 2a + 3b)[a - (2a + 3b)]

= (3a + 3b)(a - 2a - 3b)

= (3a + 3b)(-a - 3b)

= -3(a + b)(a + 3b)

Hence, a² - (2a + 3b)² = -3(a + b)(a + 3b).

Given,

   25(2a - b)² - 81b²

= [5(2a - b)]² - (9b)²

= [5(2a - b) + 9b][5(2a - b) - 9b]

= (10a - 5b + 9b)(10a - 5b - 9b)

= (10a + 4b)(10a - 14b)

= 2(5a + 2b) × 2(5a - 7b)

= 4(5a + 2b)(5a - 7b).

Hence, 25(2a - b)² - 81b² = 4(5a + 2b)(5a - 7b).

Given,

   50a³ - 2a

= 2a[25a² - 1]

= 2a[(5a)² - 1²]

= 2a(5a + 1)(5a - 1).

Hence, 50a³ - 2a = 2a(5a + 1)(5a - 1).

Given,

   4a²b - 9b³

= b[4a² - 9b²]

= b[(2a)² - (3b)²]

= b(2a + 3b)(2a - 3b).

Hence, 4a²b - 9b³ = b(2a + 3b)(2a - 3b).

Given,

   9(a - 2)² - 16(a + 2)²

= {[3(a - 2)]² - [4(a + 2)]²}

= {[3(a - 2) + 4(a + 2)][3(a - 2) - 4(a + 2)]}

= [(3a - 6 + 4a + 8)(3a - 6 - 4a - 8)]

= (7a + 2)(-a - 14)

= -(7a + 2)(a + 14).

Hence, 9(a - 2)² - 16(a + 2)² = -(7a + 2)(a + 14).

Given,

   (a + b)³ - a - b

= (a + b)³ - (a + b)

= (a + b)[(a + b)² - 1]

= (a + b)[(a + b)² - 1²]

= (a + b)(a + b + 1)(a + b - 1).

Hence, (a + b)³ - a - b = (a + b)(a + b + 1)(a + b - 1).

Given,

   a(a - 1) - b(b - 1)

= a² - a - b² + b

= a² - b² - a + b

= (a + b)(a - b) - 1(a - b)

= (a - b)(a + b - 1).

Hence, a(a - 1) - b(b - 1) = (a - b)(a + b - 1).

Given,

   4a² - (4b² + 4bc + c²)

= 4a² - [(2b)² + 2 × 2b × c + c²]

= (2a)² - (2b + c)²

= (2a + 2b + c)[2a - (2b + c)]

= (2a + 2b + c)(2a - 2b - c).

Hence, 4a² - (4b² + 4bc + c²) = (2a + 2b + c)(2a - 2b - c).

Given,

   4a² - 49b² + 2a - 7b

= (2a)² - (7b)² + 2a - 7b

= (2a + 7b)(2a - 7b) + (2a - 7b)

= (2a - 7b)(2a + 7b + 1).

Hence, 4a² - 49b² + 2a - 7b = (2a - 7b)(2a + 7b + 1).

Given,

   4a² - 12a + 9 - 49b²

= (2a)² - 2 × 2a × 3 + (3)² - (7b)²

= (2a - 3)² - (7b)²

= (2a - 3 + 7b)(2a - 3 - 7b).

Hence, 4a² - 12a + 9 - 49b² = (2a - 3 + 7b)(2a - 3 - 7b).

Given,

   4xy - x² - 4y² + z²

= -[x² + 4y² - 4xy - z²]

= -[(x)² + (2y)² - 2 × x × 2y - (z)²]

= -[(x - 2y)² - (z)²]

= (z)² - (x - 2y)²

= (z + x - 2y)(z - x + 2y).

Hence, 4xy - x² - 4y² + z² = (z + x - 2y)(z - x + 2y).

Given,

   a² + b² - c² - d² + 2ab - 2cd

= a² + b² + 2ab - c² - d² - 2cd

= (a + b)² - (c² + d² + 2cd)

= (a + b)² - (c + d)²

= (a + b + c + d)(a + b - c - d).

Hence, a² + b² - c² - d² + 2ab - 2cd = (a + b + c + d)(a + b - c - d).

Given,

   4x² - 12ax - y² - z² - 2yz + 9a²

= 4x² - 12ax + 9a² - y² - z² - 2yz

= (2x)² - 2 × 2x × 3a + (3a)² - (y² + z² + 2yz)

= (2x - 3a)² - (y + z)²

= (2x - 3a + y + z)[2x - 3a - (y + z)]

= (2x - 3a + y + z)(2x - 3a - y - z).

Hence, 4x² - 12ax - y² - z² - 2yz + 9a² = (2x - 3a + y + z)(2x - 3a - y - z).

Given,

   (a² - 1)(b² - 1) + 4ab

= a²b² - a² - b² + 1 + 4ab

= a²b² - a² - b² + 1 + 2ab + 2ab

= a²b² + 2ab + 1 - a² - b² + 2ab

= (ab + 1)² - (a² + b² - 2ab)

= (ab + 1)² - (a - b)²

= (ab + 1 + a - b)(ab + 1 - a + b).

Hence, (a² - 1)(b² - 1) + 4ab = (ab + 1 + a - b)(ab + 1 - a + b).

Given,

   x⁴ + x² + 1

Adding and subtracting x² in the polynomial,

⇒ x⁴ + x² + 1 + x² - x²

= x⁴ + 2x² + 1 - x²

= (x²)² + 2 × x² × 1 + (1)² - (x)²

= (x² + 1)² - (x)²

= (x² + 1 + x)(x² + 1 - x).

Hence, x⁴ + x² + 1 = (x² + 1 + x)(x² + 1 - x).

Given,

   (a² + b² - 4c²)² - 4a²b²

= (a² + b² - 4c²)² - (2ab)²

= (a² + b² - 4c² + 2ab)(a² + b² - 4c² - 2ab)

= (a² + b² + 2ab - 4c²)(a² + b²- 2ab - 4c²)

= [(a + b)² - (2c)²][(a - b)² - (2c)²]

= (a + b + 2c)(a + b - 2c)(a - b + 2c)(a - b - 2c).

Hence, (a² + b² - 4c²)² - 4a²b² = (a + b + 2c)(a + b - 2c)(a - b + 2c)(a - b - 2c).

Given,

   (x² + 4y² - 9z²)² - 16x²y²

= (x² + 4y² - 9z²)² - (4xy)²

= (x² + 4y² - 9z² + 4xy)(x² + 4y² - 9z² - 4xy)

= [x² + (2y)² + 4xy - 9z²][x² + (2y)² - 4xy - 9z²]

= [(x + 2y)² - (3z)²][(x - 2y)² - (3z)²]

= (x + 2y + 3z)(x + 2y - 3z)(x - 2y + 3z)(x - 2y - 3z).

Hence, (x² + 4y² - 9z²)² - 16x²y² = (x + 2y + 3z)(x + 2y - 3z)(x - 2y + 3z)(x - 2y - 3z).

Given,

   (a + b)² - a² + b²

= (a + b)² - (a² - b²)

= (a + b)² - (a + b)(a - b)

= (a + b)[(a + b) - (a - b)]

= (a + b)[a + b - a + b]

= 2b(a + b).

Hence, (a + b)² - a² + b² = 2b(a + b).

Given,

   a² - b² - (a + b)²

= (a + b)(a - b) - (a + b)²

= (a + b)[(a - b) - (a + b)]

= (a + b)[a - b - a - b]

= -2b(a + b).

Hence, a² - b² - (a + b)² = -2b(a + b).

Given,

   a³ - 8

= (a)³ - (2)³

= (a - 2)[(a)² + 2 × a + (2)²]

= (a - 2)(a² + 2a + 4)

Hence, Option 2 is the correct option.

Given,

   27 + 8x³

= (3)³ + (2x)³

= (3 + 2x)[(3)² - 3 × 2x + (2x)²]

= (3 + 2x)(9 - 6x + 4x²)

Hence, Option 3 is the correct option.

Given,

   8a³ + 1

= (2a)³ + (1)³

= (2a + 1)[(2a)² - 2a × 1 + (1)²]

= (2a + 1)(4a² - 2a + 1)

Hence, Option 1 is the correct option.

Given,

$$\begin{aligned}\phantom{=} \Big(\frac{x^3}{8} - \frac{y^3}{27}\Big) \\[1em] = \Big(\frac{x}{2}\Big)^3 - \Big(\frac{y}{3}\Big)^3 \\[1em] = \Big(\frac{x}{2} - \frac{y}{3}\Big)\Big[\Big(\frac{x}{2}\Big)^2 + \frac{x}{2} \times \frac{y}{3} + \Big(\frac{y}{3}\Big)^2\Big] \\[1em] = \Big(\frac{x}{2} - \frac{y}{3}\Big)\Big(\frac{x^2}{4} + \frac{xy}{6} + \frac{y^2}{9}\Big).\end{aligned}$$

Hence, Option 3 is the correct option.

Factorise :

Given,

   64 - a³b³

= (4)³ - (ab)³

= (4 - ab)[(4)² + 4 × ab + (ab)²]

= (4 - ab)(16 + 4ab + a²b²).

Hence, 64 - a³b³ = (4 - ab)(16 + 4ab + a²b²).

Given,

   a⁶ + 27b³

= (a²)³ + (3b)³

= (a² + 3b)[(a²)² - a² × 3b + (3b)²]

= (a² + 3b)(a⁴ - 3a²b + 9b²).

Hence, a⁶ + 27b³ = (a² + 3b)(a⁴ - 3a²b + 9b²).

Given,

   3x⁷y - 81x⁴y⁴

= 3x⁴y[x³ - 27y³]

= 3x⁴y[(x)³ - (3y)³]

= 3x⁴y(x - 3y)[(x)² + x × 3y + (3y)²]

= 3x⁴y(x - 3y)(x² + 3xy + 9y²).

Hence, 3x⁷y - 81x⁴y⁴ = 3x⁴y(x - 3y)(x² + 3xy + 9y²).

Given,

$$\begin{aligned}\phantom{=} a^3 - \frac{27}{a^3} \\[1em] = (a)^3 - \Big(\frac{3}{a}\Big)^3 \\[1em] = \Big(a - \frac{3}{a}\Big)\Big[(a)^2 + a \times \frac{3}{a} + \Big(\frac{3}{a}\Big)^2\Big] \\[1em] = \Big(a - \frac{3}{a}\Big)\Big(a^2 + 3 + \frac{9}{a^2}\Big).\end{aligned}$$

Hence, $a^3 - \frac{27}{a^3} = \Big(a - \frac{3}{a}\Big)\Big(a^2 + 3 + \frac{9}{a^2}\Big).$

Given,

    a³ + 0.064

= (a)³ + (0.4)³

= (a + 0.4)[a² - 0.4 × a + (0.4)²]

= (a + 0.4)(a² - 0.4a + 0.16)

Hence, a³ + 0.064 = (a + 0.4)(a² - 0.4a + 0.16)

Given,

    (x - y)³ - 8x³

= (x - y)³ - (2x)³

= (x - y - 2x)[(x - y)² + 2x(x - y) + (2x)²]

= (-x - y)[x² + y² - 2xy + 2x² - 2xy + 4x²]

= -(x + y)(7x² - 4xy + y²).

Hence, (x - y)³ - 8x³ = -(x + y)(7x² - 4xy + y²).

Given,

$$\begin{aligned}\phantom{=}\Big(\frac{8a^3}{27} - \frac{b^3}{8}\Big) \\[1em] = \Big(\frac{2a}{3}\Big)^3 - \Big(\frac{b}{2}\Big)^3 \\[1em] = \Big(\frac{2a}{3} - \frac{b}{2}\Big)\Big[\Big(\frac{2a}{3}\Big)^2 + \frac{2a}{3} \times \frac{b}{2} + \Big(\frac{b}{2}\Big)^2\Big] \\[1em] = \Big(\frac{2a}{3} - \frac{b}{2}\Big)\Big(\frac{4a^2}{9} + \frac{ab}{3} + \frac{b^2}{4}\Big) \\[1em]\end{aligned}$$

Hence, $\Big(\frac{8a^3}{27} - \frac{b^3}{8}\Big) = \Big(\frac{2a}{3} - \frac{b}{2}\Big)\Big(\frac{4a^2}{9} + \frac{ab}{3} + \frac{b^2}{4}\Big).$

Given,

    a⁶ - b⁶

= (a³)² - (b³)²

= (a³ + b³)(a³ - b³)

= (a + b)(a² - ab + b²)(a - b)(a² + ab + b²)

= (a + b)(a - b)(a² - ab + b²)(a² + ab + b²).

Hence, a⁶ - b⁶ = (a + b)(a - b)(a² - ab + b²)(a² + ab + b²).

Given,

    a⁶ - 7a³ - 8

= (a³)² - 7a³ - 8

Substituting a³ = x, we get :

= x² - 7x - 8

= x² - 8x + x - 8

= x(x - 8) + 1(x - 8)

= (x - 8)(x + 1)

= (a³ - 8)(a³ + 1)

= (a³ - 2³)(a³ + 1³)

= (a - 2)(a² + a × 2 + 2²)(a + 1)(a² - a × 1 + 1²)

= (a - 2)(a² + 2a + 4)(a + 1)(a² - a + 1)

= (a - 2)(a + 1)(a² + 2a + 4)(a² - a + 1).

Hence, a⁶ - 7a³ - 8 = (a - 2)(a + 1)(a² + 2a + 4)(a² - a + 1).

Given,

    a³ - 27b³ + 2a²b - 6ab²

= (a)³ - (3b)³ + 2ab(a - 3b)

= (a - 3b)[(a)² + a × 3b + (3b)²] + 2ab(a - 3b)

= (a - 3b)[a² + 3ab + 9b²] + 2ab(a - 3b)

= (a - 3b)(a² + 3ab + 9b² + 2ab)

= (a - 3b)(a² + 5ab + 9b²).

Hence, a³ - 27b³ + 2a²b - 6ab² = (a - 3b)(a² + 5ab + 9b²).

Given,

    8a³ - b³ - 4ax + 2bx

= (2a)³ - (b)³ - 2x(2a - b)

= (2a - b)[(2a)² + 2a × b + (b)²] - 2x(2a - b)

= (2a - b)(4a² + 2ab + b²) - 2x(2a - b)

= (2a - b)(4a² + 2ab + b² - 2x).

Hence, 8a³ - b³ - 4ax + 2bx = (2a - b)(4a² + 2ab + b² - 2x).

Given,

    a - b - a³ + b³

= (a - b) - (a³ - b³)

= (a - b) - (a - b)(a² + ab + b²)

= (a - b)(1 - a² - ab - b²).

Hence, a - b - a³ + b³ = (a - b)(1 - a² - ab - b²).

Given,

$$\begin{aligned}\phantom{=}x^4 + \frac{1}{x^4} - 2 \\[1em] = (x^2)^2 + \Big(\frac{1}{x^2}\Big)^2 - 2 \times x^2 \times \frac{1}{x^2} \\[1em] = \Big(x^2 - \frac{1}{x^2}\Big)^2 \\[1em] = \Big[\Big(x - \frac{1}{x}\Big)\Big(x + \frac{1}{x}\Big)\Big]^2 \\[1em] = \Big(x - \frac{1}{x}\Big)^2\Big(x + \frac{1}{x}\Big)^2.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\phantom{=}x^2 + \frac{1}{x^2} - 3 \\[1em] = x^2 + \frac{1}{x^2} - 2 - 1 \\[1em] = x^2 + \frac{1}{x^2} - \Big(2 \times x \times \frac{1}{x}\Big) - 1 \\[1em] = \Big[x^2 + \frac{1}{x^2} - \Big(2 \times x \times \frac{1}{x}\Big)\Big] - 1 \\[1em] = \Big(x - \frac{1}{x}\Big)^2 - 1 \\[1em] = \Big(x - \frac{1}{x}\Big)^2 - 1^2 \\[1em] = \Big(x - \frac{1}{x} + 1\Big)\Big(x - \frac{1}{x} - 1\Big).\end{aligned}$$

Hence, Option 4 is the correct option.

Factorise :

Given,

$$\begin{aligned}\phantom{=} x^2 + \frac{1}{4x^2} + 1 - 7x - \frac{7}{2x} \\[1em] = x^2 + \Big(\frac{1}{2x}\Big)^2 + 1 - 7\Big(x + \frac{1}{2x}\Big) \\[1em] = x^2 + \Big(\frac{1}{2x}\Big)^2 + 2 \times x \times \frac{1}{2x} - 7\Big(x + \frac{1}{2x}\Big) \\[1em] = \Big(x + \frac{1}{2x}\Big)^2 - 7\Big(x + \frac{1}{2x}\Big) \\[1em] = \Big(x + \frac{1}{2x}\Big)\Big(x + \frac{1}{2x} - 7\Big).\end{aligned}$$

Hence, $x^2 + \frac{1}{4x^2} + 1 - 7x - \frac{7}{2x} = \Big(x + \frac{1}{2x}\Big)\Big(x + \frac{1}{2x} - 7\Big).$

Given,

$$\begin{aligned}\phantom{=} 9a^2 + \frac{1}{9a^2} - 2 - 12a + \frac{4}{3a} \\[1em] = (3a)^2 + \Big(\frac{1}{3a}\Big)^2 - 2 - 4\Big(3a - \frac{1}{3a}\Big) \\[1em] = \Big(3a - \frac{1}{3a}\Big)^2 - 4\Big(3a - \frac{1}{3a}\Big) \\[1em] = \Big(3a - \frac{1}{3a}\Big)\Big(3a - \frac{1}{3a} - 4\Big).\end{aligned}$$

Hence, $9a^2 + \frac{1}{9a^2} - 2 - 12a + \frac{4}{3a} = \Big(3a - \frac{1}{3a}\Big)\Big(3a - \frac{1}{3a} - 4\Big).$

Given,

$$\begin{aligned}\phantom{=} x^2 + \frac{a^2 + 1}{a}x + 1 \\[1em] = \frac{ax^2 + (a^2 + 1)x + a}{a} \\[1em] = \frac{ax^2 + a^2x + x + a}{a} \\[1em] = \frac{ax(x + a) + (x + a)}{a} \\[1em] = \frac{(x + a)(ax + 1)}{a} \\[1em] = (x + a)\frac{ax + 1}{a} \\[1em] = (x + a)\Big(x + \frac{1}{a}\Big).\end{aligned}$$

Hence, $x^2 + \frac{a^2 + 1}{a}x + 1 = (x + a)\Big(x + \frac{1}{a}\Big).$

Given,

    x⁴ + y⁴ - 27x²y²

= x⁴ + y⁴ - 2x²y² - 25x²y²

= (x² - y²)² - (5xy)²

= (x² - y² + 5xy)(x² - y² - 5xy).

Hence, x⁴ + y⁴ - 27x²y² = (x² - y² + 5xy)(x² - y² - 5xy).

Given,

    4x⁴ + 9y⁴ + 11x²y²

= (2x²)² + (3y²)² + 12x²y² - x²y²

= (2x²)² + (3y²)² + 2 × 2x² × 3y² - x²y²

= (2x² + 3y²)² - (xy)²

= (2x² + 3y² + xy)(2x² + 3y² - xy).

Hence, 4x⁴ + 9y⁴ + 11x²y² = (2x² + 3y² + xy)(2x² + 3y² - xy).

Given,

$$\begin{aligned}\phantom{=}x^2 + \frac{1}{x^2} - 3 \\[1em] = x^2 + \frac{1}{x^2} - 2 - 1 \\[1em] = \Big(x - \frac{1}{x}\Big)^2 - 1^2 \\[1em] = \Big(x - \frac{1}{x} + 1\Big)\Big(x - \frac{1}{x} - 1\Big).\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2} - 3 = \Big(x - \frac{1}{x} + 1\Big)\Big(x - \frac{1}{x} - 1\Big).$

Given,

    a - b - 4a² + 4b²

= a - b - 4(a² - b²)

= (a - b) - 4(a + b)(a - b)

= (a - b)[1 - 4(a + b)]

= (a - b)(1 - 4a - 4b).

Hence, a - b - 4a² + 4b² = (a - b)(1 - 4a - 4b).

Given,

    (2a - 3)² - 2(2a - 3)(a - 1) + (a - 1)²

= [(2a - 3) - (a - 1)]²

= (2a - 3 - a + 1)²

= (a - 2)².

Hence, (2a - 3)² - 2(2a - 3)(a - 1) + (a - 1)² = (a - 2)².

Given,

    (a² - 3a)(a² - 3a + 7) + 10

Substituting a² - 3a = x, we get :

⇒ x(x + 7) + 10

= x² + 7x + 10

= x² + 2x + 5x + 10

= x(x + 2) + 5(x + 2)

= (x + 5)(x + 2)

= (a² - 3a + 5)(a² - 3a + 2)

= (a² - 3a + 5)(a² - 2a - a + 2)

= (a² - 3a + 5)[a(a - 2) -1(a - 2)]

= (a² - 3a + 5)(a - 1)(a - 2)

Hence, (a² - 3a)(a² - 3a + 7) + 10 = (a² - 3a + 5)(a - 1)(a - 2).

Given,

    (a² - a)(4a² - 4a - 5) - 6

= (a² - a)[4(a² - a) - 5] - 6

Substituting a² - a = x, we get :

= x(4x - 5) - 6

= 4x² - 5x - 6

= 4x² - 8x + 3x - 6

= 4x(x - 2) + 3(x - 2)

= (x - 2)(4x + 3)

= (a² - a - 2)[4(a² - a) + 3]

= (a² - a - 2)(4a² - 4a + 3)

= (a² - 2a + a - 2)(4a² - 4a + 3)

= [a(a - 2) + 1(a - 2)](4a² - 4a + 3)

= (a - 2)(a + 1)(4a² - 4a + 3).

Hence, (a² - a)(4a² - 4a - 5) - 6 = (a - 2)(a + 1)(4a² - 4a + 3).

Given,

    x⁴ + y⁴ - 3x²y²

= (x²)² + (y²)² - 2x²y² - x²y²

= (x² - y²)² - (xy)²

= (x² - y² + xy)(x² - y² - xy).

Hence, x⁴ + y⁴ - 3x²y² = (x² - y² + xy)(x² - y² - xy).

Given,

    5a² - b² - 4ab + 7a - 7b

= 5a² - 4ab - b² + 7a - 7b

= 5a² - 5ab + ab - b² + 7a - 7b

= 5a(a - b) + b(a - b) + 7(a - b)

= (a - b)(5a + b + 7).

Hence, 5a² - b² - 4ab + 7a - 7b = (a - b)(5a + b + 7).

Given,

    12(3x - 2y)² - 3x + 2y - 1

= 12(3x - 2y)² - (3x - 2y) - 1

Substituting (3x - 2y) = a, we get :

= 12a² - a - 1

= 12a² - 4a + 3a - 1

= 4a(3a - 1) + 1(3a - 1)

= (4a + 1)(3a - 1)

= [4(3x - 2y) + 1][3(3x - 2y) - 1]

= (12x - 8y + 1)(9x - 6y - 1).

Hence, 12(3x - 2y)² - 3x + 2y - 1 = (12x - 8y + 1)(9x - 6y - 1).

Given,

    4(2x - 3y)² - 8x + 12y - 3

= 4(2x - 3y)² - 4(2x - 3y) - 3

Substituting (2x - 3y) = a, we get :

= 4a² - 4a - 3

= 4a² - 6a + 2a - 3

= 2a(2a - 3) + 1(2a - 3)

= (2a + 1)(2a - 3)

= [2(2x - 3y) + 1][2(2x - 3y) - 3]

= (4x - 6y + 1)(4x - 6y - 3).

Hence, 4(2x - 3y)² - 8x + 12y - 3 = (4x - 6y + 1)(4x - 6y - 3).

Given,

    3 - 5x + 5y - 12(x - y)²

= 3 - 5(x - y) - 12(x - y)²

Substituting x - y = a, we get :

= 3 - 5a - 12a²

= 3 - 9a + 4a - 12a²

= 3(1 - 3a) + 4a(1 - 3a)

= (1 - 3a)(3 + 4a)

= [1 - 3(x - y)][3 + 4(x - y)]

= (1 - 3x + 3y)(3 + 4x - 4y).

Hence, 3 - 5x + 5y - 12(x - y)² = (1 - 3x + 3y)(3 + 4x - 4y).

Given,

$$\begin{aligned}\Rightarrow \frac{(a + b)(a^2 - ab + b^2)}{(a^2b + ab^2)}\\[1em] \Rightarrow \frac{(a + b)(a^2 - ab + b^2)}{ab(a + b)}\\[1em] \Rightarrow \frac{(a^2 - ab + b^2)}{ab}\\[1em] \Rightarrow \frac{a^2}{ab} - \frac{ab}{ab} + \frac{b^2}{ab} \\[1em] \Rightarrow \frac{a}{b} - 1 + \frac{b}{a}\end{aligned}$$

Hence, option 1 is the correct option.

Given, x² + 3x + 2 and x² - 2x - 3

The factors of x² + 3x + 2

⇒ x² + 2x + x + 2

⇒ x(x + 2) + 1(x + 2)

⇒ (x + 2)(x + 1).

The factors of x² - 2x - 3

⇒ x² - 3x + x - 3

⇒ x(x - 3) + 1(x - 3)

⇒ (x - 3)(x + 1)

L.C.M. = (x + 1)(x + 2)(x - 3)

Hence, option 3 is the correct option.

Given, x² + 3x + 2 and x² - 2x - 3

The factors of x² + 3x + 2

⇒ x² + 2x + x + 2

⇒ x(x + 2) + 1(x + 2)

⇒ (x + 2)(x + 1)

The factors of x² - 2x - 3

⇒ x² - 3x + x - 3

⇒ x(x - 3) + 1(x - 3)

⇒ (x - 3)(x + 1)

H.C.F. = (x + 1)

Hence, option 1 is the correct option.

Solving,

⇒ (3a - 1)² - 6a + 2

⇒ (3a)² + 1² - 2 x 3a x 1 - 6a + 2

⇒ 9a² + 1 - 6a - 6a + 2

⇒ 9a² - 12a + 3

⇒ 3(3a² - 4a + 1)

⇒ 3(3a² - 3a - a + 1)

⇒ 3[3a(a - 1) - 1(a - 1)]

⇒ 3(3a - 1)(a - 1).

Hence, option 2 is the correct option.

Given,

$$\begin{aligned}\frac{1}{3} x^2 - 2x - 9\\[1em] \Rightarrow \frac{1}{3} x^2 - \frac{6}{3}x - \frac{27}{3}\\[1em] \Rightarrow \frac{1}{3} (x^2 - 6x - 27)\\[1em] \Rightarrow \frac{1}{3} (x^2 - 9x + 3x - 27)\\[1em] \Rightarrow \frac{1}{3} [x(x - 9) + 3(x - 9)]\\[1em] \Rightarrow \frac{1}{3} (x - 9)(x + 3)\end{aligned}$$

Hence, option 1 is the correct option.

Given, total number of men = (x² + 3x)

Total number of days = (x² - 2x)

Total work = (x² + 3x)(x² - 2x)

One day work of 1 man = $\frac{1}{(x^2 + 3x)(x^2 - 2x)}$

Hence, option 4 is the correct option.

Given, cost of each article = ₹ (x - 1)

Total cost = ₹ (x³ - x)

$$\begin{aligned}\Rightarrow \text{No. of articles bought} = \frac{\text{Total cost}}{\text{Cost of each article}} \\[1em] = \frac{x^3 - x}{x - 1}\\[1em] = \frac{x(x^2 - 1)}{x - 1}\\[1em] = \frac{x(x^2 - 1^2)}{x - 1}\\[1em] = \frac{x(x - 1)(x + 1)}{x - 1}\\[1em] = x(x + 1)\\[1em] = x^2 + x.\end{aligned}$$

∴ Both the statements are true.

Hence, option 1 is the correct option.

Give, Area of rectangle = x² - 5x + 6

Longer side = (x - 2)

Factorise the area,

⇒ x² - 5x + 6 = 0

⇒ x² - 3x - 2x + 6 = 0

⇒ x(x - 3) - 2(x - 3) = 0

⇒ (x - 2) (x - 3) = 0

So, the given two sides of rectangle are:

(x - 2) (x - 3)

Since x > 3,

∴ (x - 2) > (x - 3)

So, (x - 2) is the longer side.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Given,

Distance = (x² - 7x + 12) km

Time = (x² - 16) hrs

By formula,

Speed = $\frac{\text{Distance}}{\text{Time}}$

= $\frac{x^2 - 7x + 12}{x^2 - 16}$ km/hr

∴ A is true, but R is false.

Hence, option 1 is the correct option.

Factorise :

Given,

$$\begin{aligned}\phantom{=}a^2 + \frac{1}{a^2} - 2 - 3a + \frac{3}{a} \\[1em] = \Big(a - \frac{1}{a}\Big)^2 - 3\Big(a - \frac{1}{a}\Big) \\[1em] = \Big(a - \frac{1}{a}\Big)\Big(a - \frac{1}{a} - 3\Big).\end{aligned}$$

Hence, $a^2 + \frac{1}{a^2} - 2 - 3a + \frac{3}{a} = \Big(a - \frac{1}{a}\Big)\Big(a - \frac{1}{a} - 3\Big).$

Given,

    x² + y² + x + y + 2xy

= x² + y² + 2xy + x + y

= (x + y)² + (x + y)

= (x + y)(x + y + 1).

Hence, x² + y² + x + y + 2xy = (x + y)(x + y + 1).

Given,

    a² + 4b² - 3a + 6b - 4ab

= a² + 4b² - 4ab - 3a + 6b

= a² + (2b)² - 2 × a × 2b - 3a + 6b

= (a - 2b)² - 3(a - 2b)

= (a - 2b)(a - 2b - 3).

Hence, a² + 4b² - 3a + 6b - 4ab =(a - 2b)(a - 2b - 3).

Given,

    m(x - 3y)² + n(3y - x) + 5x - 15y

= m(x - 3y)² - n(x - 3y) + 5(x - 3y)

= (x - 3y)[m(x - 3y) - n + 5]

= (x - 3y)(mx - 3my - n + 5).

Hence, m(x - 3y)² + n(3y - x) + 5x - 15y = (x - 3y)(mx - 3my - n + 5).

Given,

    x(6x - 5y) - 4(6x - 5y)²

= (6x - 5y)[x - 4(6x - 5y)]

= (6x - 5y)(x - 24x + 20y)

= (6x - 5y)(20y - 23x).

Hence, x(6x - 5y) - 4(6x - 5y)² = (6x - 5y)(20y - 23x).

Given,

$$\begin{aligned}\phantom{=}\frac{1}{35} + \frac{12}{35}a + a^2 \\[1em] = a^2 + \frac{12}{35}a + \frac{1}{35} \\[1em] = \frac{35a^2 + 12a + 1}{35} \\[1em] = \frac{35a^2 + 5a + 7a + 1}{35} \\[1em] = \frac{5a(7a + 1) + 1(7a + 1)}{35} \\[1em] = \frac{(7a + 1)(5a + 1)}{35}.\end{aligned}$$

Hence, $\frac{1}{35} + \frac{12}{35}a + a^2 = \frac{(7a + 1)(5a + 1)}{35}$.

Given,

    (x² - 3x)(x² - 3x - 1) - 20

Substituting x² - 3x = a, we get :

⇒ a(a - 1) - 20

= a² - a - 20

= a² - 5a + 4a - 20

= a(a - 5) + 4(a - 5)

= (a - 5)(a + 4)

= (x² - 3x - 5)(x² - 3x + 4).

Hence, (x² - 3x)(x² - 3x - 1) - 20 = (x² - 3x - 5)(x² - 3x + 4).

(i) Given,

    x² - 3x - 54

= x² - 9x + 6x - 54

= x(x - 9) + 6(x - 9)

= (x - 9)(x + 6).

Hence, the above equation is factorisable and x² - 3x - 54 = (x - 9)(x + 6).

(ii) Given,

    2x² - 7x - 15

= 2x² - 10x + 3x - 15

= 2x(x - 5) + 3(x - 5)

= (2x + 3)(x - 5).

Hence, the above equation is factorisable and 2x² - 7x - 15 = (2x + 3)(x - 5).

(iii) Given,

    2x² + 2x - 75

Hence, the above equation is not factorisable.

(iv) Given,

    3x² + 4x - 10

Hence, the above equation is not factorisable.

(v) Given,

    x(2x - 1) - 1

= 2x² - x - 1

= 2x² - 2x + x - 1

= 2x(x - 1) + 1(x - 1)

= (x - 1)(2x + 1).

Hence, the above equation is factorisable and x(2x - 1) - 1 = (x - 1)(2x + 1).

(i) Given,

$$\begin{aligned}\phantom{=}4\sqrt{3}x^2 + 5x - 2\sqrt{3} \\[1em] = 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3}\\[1em] = 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) \\[1em] = (\sqrt{3}x + 2)(4x - \sqrt{3}).\end{aligned}$$

Hence, $4\sqrt{3}x^2 + 5x - 2\sqrt{3} = (\sqrt{3}x + 2)(4x - \sqrt{3}).$

(ii) Given,

$$\begin{aligned}\phantom{=}7\sqrt{2}x^2 - 10x - 4\sqrt{2} \\[1em] = 7\sqrt{2}x^2 - 14x + 4x - 4\sqrt{2}\\[1em] = 7\sqrt{2}x(x - \sqrt{2}) + 4(x - \sqrt{2}) \\[1em] = (x - \sqrt{2})(7\sqrt{2}x + 4).\end{aligned}$$

Hence, $7\sqrt{2}x^2 - 10x - 4\sqrt{2} = (x - \sqrt{2})(7\sqrt{2}x + 4).$

Given,

Area = 12x² - 35x + 25

⇒ lb = 12x² - 35x + 25

⇒ lb = 12x² - 15x - 20x + 25

⇒ lb = 3x(4x - 5) - 5(4x - 5)

⇒ lb = (4x - 5)(3x - 5).

Hence, if length = (4x - 5) then breadth = (3x - 5) and if length = (3x - 5) then breadth = (4x - 5).

Factorise :

Given,

    9a² - (a² - 4)²

= (3a)² - (a² - 4)²

= (3a + a² - 4)[3a - (a² - 4)]

= (a² + 3a - 4)(4 - a² + 3a)

= (a² + 4a - a - 4).-(a² - 3a - 4)

= [a(a + 4) - 1(a + 4)].-[a² - 4a + a - 4]

= (a + 4)(a - 1).-[a(a - 4) + 1(a - 4)]

= (a + 4)(a - 1).-(a - 4)(a + 1)

= (a + 4)(a - 1)(4 - a)(a + 1).

Hence, 9a² - (a² - 4)² = (a + 4)(a - 1)(4 - a)(a + 1).

Given,

$$\begin{aligned}\phantom{=} x^2 + \frac{1}{x^2} - 11 \\[1em] = x^2 + \frac{1}{x^2} - 2 - 9 \\[1em] = \Big(x - \frac{1}{x}\Big)^2 - 9 \\[1em] = \Big(x - \frac{1}{x}\Big)^2 - 3^2 \\[1em] = \Big(x - \frac{1}{x} + 3\Big)\Big(x - \frac{1}{x} - 3\Big).\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2} - 11 = \Big(x - \frac{1}{x} + 3\Big)\Big(x - \frac{1}{x} - 3\Big).$

Given,

$$\begin{aligned}\phantom{=} 4x^2 + \frac{1}{4x^2} + 1 \\[1em] = 4x^2 + \frac{1}{4x^2} + 2 - 1 \\[1em] = (2x)^2 + \Big(\frac{1}{2x}\Big)^2 + 2 \times 2x \times \frac{1}{2x} - 1 \\[1em] = \Big(2x + \frac{1}{2x}\Big)^2 - 1^2 \\[1em] = \Big(2x + \frac{1}{2x} + 1\Big)\Big(2x + \frac{1}{2x} - 1\Big).\end{aligned}$$

Hence, $4x^2 + \frac{1}{4x^2} + 1 = \Big(2x + \frac{1}{2x} + 1\Big)\Big(2x + \frac{1}{2x} - 1\Big).$

Given,

    4x⁴ - x² - 12x - 36

= 4x⁴ - [x² + 12x + 36]

= 4x⁴ - [x² + 6x + 6x + 36]

= 4x⁴ - [x(x + 6) + 6(x + 6)]

= 4x⁴ - (x + 6)(x + 6)

= (2x²)² - (x + 6)²

= (2x² + x + 6)(2x² - x - 6).

= (2x² + x + 6)(2x² - 4x + 3x - 6)

= (2x² + x + 6)[2x(x - 2) + 3(x - 2)]

= (2x² + x + 6)(x - 2)(2x + 3).

Hence, 4x⁴ - x² - 12x - 36 = (2x² + x + 6)(x - 2)(2x + 3).

Given,

    a²(b + c) - (b + c)³

= (b + c)[a² - (b + c)²]

= (b + c)(a + b + c)[a - (b + c)]

= (b + c)(a + b + c)(a - b - c).

Hence, a²(b + c) - (b + c)³ = (b + c)(a + b + c)(a - b - c).

Given,

    2x³ + 54y³ - 4x - 12y

= 2(x³ + 27y³) - 4(x + 3y)

= 2[(x)³ + (3y)³] - 4(x + 3y)

= 2(x + 3y)(x² - 3xy + 9y²) - 4(x + 3y) [∵ a³ + b³ = (a + b)(a² - ab + b²)]

= 2(x + 3y)(x² - 3xy + 9y² - 2).

Hence, 2x³ + 54y³ - 4x - 12y = 2(x + 3y)(x² - 3xy + 9y² - 2).

Given,

    1029 - 3x³

= 3(343 - x³)

= 3[(7)³ - (x)³]

= 3(7 - x)[(7)² + 7x + x²] [∵ a³ - b³ = (a - b)(a² + ab + b²)]

= 3(7 - x)(x² + 7x + 49).

Hence, 1029 - 3x³ = 3(7 - x)(x² + 7x + 49).

(i) We know that

a³ - b³ = (a - b)(a² + ab + b²)

Factorising 13³ - 5³, we get :

⇒ 13³ - 5³ = (13 - 5)[13² + 13 × 5 + 5²]

= 8(169 + 65 + 25)

= 8 × 259, which is divisible by 8.

Hence, proved that 13³ - 5³ is divisible by 8.

(ii) We know that

a³ + b³ = (a + b)(a² - ab + b²)

Factorising 35³ + 27³, we get :

⇒ 35³ + 27³ = (35 + 27)[(35)² - 35 × 27 + (27)²]

= 62[1225 - 945 + 729]

= 62 × 1009, which is divisible by 62.

Hence, proved that 35³ + 27³ is divisible by 62.

Substituting a = 5.67 and b = 4.33, we get :

$$\begin{aligned}\Rightarrow \frac{a \times a \times a + b \times b \times b}{a \times a - a \times b + b \times b} \\[1em] \Rightarrow \frac{a^3 + b^3}{a^2 - ab + b^2} \\[1em] \Rightarrow \frac{(a + b)(a^2 - ab + b^2)}{a^2 - ab + b^2} \\[1em] \Rightarrow (a + b) \\[1em] \Rightarrow (5.67 + 4.33) \\[1em] \Rightarrow 10.\end{aligned}$$

Hence, $\frac{5.67 \times 5.67 \times 5.67 + 4.33 \times 4.33 \times 4.33}{5.67 \times 5.67 - 5.67 \times 4.33 + 4.33 \times 4.33} = 10$.

Given,

    9x² + 3x - 8y - 64y²

= 9x² - 64y² + 3x - 8y

= (3x)² - (8y)² + 3x - 8y

= (3x + 8y)(3x - 8y) + (3x - 8y)

= (3x - 8y)(3x + 8y + 1).

Hence, 9x² + 3x - 8y - 64y² = (3x - 8y)(3x + 8y + 1).

Given,

$$\begin{aligned}\phantom{=} 2\sqrt{3}x^2 + x - 5\sqrt{3} \\[1em] = 2\sqrt{3}x^2 + 6x - 5x - 5\sqrt{3} \\[1em] = 2\sqrt{3}x(x + \sqrt{3}) - 5(x + \sqrt{3}) \\[1em] = (x + \sqrt{3})(2\sqrt{3}x - 5).\end{aligned}$$

Hence, $2\sqrt{3}x^2 + x - 5\sqrt{3} = (x + \sqrt{3})(2\sqrt{3}x - 5).$

Given,

$$\begin{aligned}\phantom{=}\frac{1}{4}(a + b)^2 - \frac{9}{16}(2a - b)^2 \\[1em] = \Big[\frac{1}{2}(a + b)\Big]^2 - \Big[\frac{3}{4}(2a - b)\Big]^2 \\[1em] = \Big[\frac{1}{2}(a + b) + \frac{3}{4}(2a - b)\Big]\Big[\frac{1}{2}(a + b) - \frac{3}{4}(2a - b)\Big] \\[1em] = \Big[\frac{2(a + b) + 3(2a - b)}{4}\Big]\Big[\frac{2(a + b) - 3(2a - b)}{4}\Big] \\[1em] = \Big(\frac{2a + 2b + 6a - 3b}{4}\Big)\Big(\frac{2a + 2b - 6a + 3b}{4}\Big)\\[1em] = \frac{8a - b}{4} \times \frac{5b - 4a}{4} \\[1em] = \frac{1}{16}(8a - b)(5b - 4a).\end{aligned}$$

Hence, $\frac{1}{4}(a + b)^2 - \frac{9}{16}(2a - b)^2 = \frac{1}{16}(8a - b)(5b - 4a).$

Given,

    2(ab + cd) - a² - b² + c² + d²

= 2ab + 2cd - a² - b² + c² + d²

= c² + d² + 2cd - (a² + b² - 2ab)

= (c + d)² - (a - b)²

= (c + d + a - b)[c + d - (a - b)]

= (c + d + a - b)(c + d - a + b).

Hence, 2(ab + cd) - a² - b² + c² + d² = (c + d + a - b)(c + d - a + b).

Given,

a² + 5a + (3 - b)(2 + b)

= a² + 5a + 6 + 3b - 2b - b²

= a² + 5a + 6 + b - b²

= (a + b + 2)(a - b + 3).

Hence, a² + 5a + (3 - b) (2 + b) = (a + b + 2) (a - b + 3).

Given,

$\frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y} + \frac{x}{z} + \frac{z}{x} + 3$

Multiplying the given expression by xyz,

= xyz × $\Big(\frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y} + \frac{x}{z} + \frac{z}{x} + 3)$

= x²z + y²z + y²x + z²x + x²y + z²y + 3xyz

= x²y + xy² + y²z + yz² + z²x + zx² + 3xyz

= (x + y + z)(xy + yz + zx)

Divide by xyz

= $\frac{(x + y + z)(xy + yz + zx)}{xyz}$

= $(x + y + z) \times \frac{xy + yz + zx}{xyz}$

= (x + y + z) $\Big(\frac{1}{z} + \frac{1}{x} + \frac{1}{y}\Big)$

Hence, $\frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y} + \frac{x}{z} + \frac{z}{x} + 3 = (x + y + z)\Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\Big)$.