Question

(b) Without finding the cubes, factorise (x – y)^3 + (y – z)^3 + (z – x)^3.

Answer 1

Answer:

ANSWER

Let a=x−2y,b=2y−3z,c=3z−x

a+b+c=x−2y+2y−3z+3z−x=0

As a+b+c=0 ⟹a

3

+b

3

+c

3

=3abc

⟹(x−2y)

3

+(2y−3z)

3

+(3z−x)

3

=3(x−2y)(2y−3z)(3z−x)