Question

Without finding cubes, factorise : (x − y) 3 + (y − z) 3 + (z − x) 3

Answer 1

Answer:

a^3 + b^2 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)

If,

a + b + c = 0

Then,

a^3 + b^3 + c^3 = 3abc

Here,

a = x - 2y

b = 2y - 3z

c = 3z - x

Thus,

(x - 2y) + (2y - 3z) + (3z - x) = 0

And then,

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