Question

17. Factorise (2x – 3y)³ + (3y – 4z)³ + (4z – 2x)³

Answer 1

Answer:

3(2x - 3y)(3y - 4z)(4z - 2x)

Note:

★ (a + b)² = a² + b² + 2ab

★ (a - b)² = a² + b² - 2ab

★ (a + b)(a - b) = a² - b²

★ (a + b)³ = a³ + b³ + 3ab(a + b)

★ (a - b)³ = a³ - b³ - 3ab(a - b)

★ a³ + b³ = (a + b)(a² + b² - ab)

★ a³ - b³ = (a - b)(a² + b² + ab)

★ a³ + b³ + c³ - 3abc

= (a + b + c)(a² + b² + c³ - ab - bc - ca)

★ If a + b + c = 0 , then ;

a³ + b³ + c³ = 3abc

Solution:

We need to factorize :

(2x – 3y)³ + (3y – 4z)³ + (4z – 2x)³

Let ,

a = 2x - 3y

b = 3y - 4z

c = 4z - 2x

Clearly,

a + b + c = 2x - 3y + 3y - 4z + 4z - 2x

a + b + c = 0

Hence,

=> a³ + b³ + c³ = 3abc

=> (2x – 3y)³ + (3y – 4z)³ + (4z – 2x)³

= 3(2x - 3y)(3y - 4z)(4z - 2x)

Hence,

Required answer is :

3(2x - 3y)(3y - 4z)(4z - 2x)