Question

Factorise: 8(x+y) 3 -27(x-y) 3

Answer 1

8 (x + y)^3 - 27 (x - y)^3

This can be rewritten as (2*(x + y))^3 - (3(x - y))^3.

This is of the form a^3 - b^3 and the factorization of a^3 - b^3 is as given below.

a^3 - b^3 = (a-b)* (a^2 + ab + b^2)

Where a = 2(x + y) and b = 3(x - y).

Substituting these in the above equation, we get

[2(x + y) - 3(x - y)] [(2(x + y))^2 + 2(x + y) * 3(x - y) + (3(x - y))^2]

For sake of simplicity, let us call the first part of the above expression A and the second part of the expression B.

Then, A = 2(x + y) - 3(x - y) and
B = 2(x + y))^2 + 2(x + y) * 3(x - y) + (3(x - y))^2

Expanding A, we get A = 2x + 2y - 3x + 3y = 5y - x

Expanding B, we get

B = 4x^2+8xy+4y^2+6x^2-6y^2+9x^2-18xy+9y^2
   = 4x^2+6x^2+9x^2+8xy-18xy+4y^2-6y^2+9y^2
   = 19x^2-10xy+7y^2

Therefore,
8 (x + y)^3 - 27 (x - y)^3 = (5y-x)*(19x^2-10xy+7y^2)

Answer 2

let   a = 2 (x+y)          b = 3 (x-y)

given  a^3 - b^3 = (a -b) (a^2 + ab +b^2)
 
Substitute values of a and b.   so factors are :
 
  [2x + 2y -3x +3y] [ 4x^2 +4y^2+8xy + 6x^2 - 6y^2 + 9 x^2 + 9 y^2 - 18 xy ]

= [ 5y - x] * [ 19 x^2 +  7 y^2 - 10 xy ]