Question

without finding the cubes, factorise (x − y)^3 + (y − z)^3 + (z − x)^3

Answer 1

when a+b+c = 0 we know that a^3+b^3+c^3 = 3 ABC

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

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

= 3 (xy-xz-y^2+yz)(z-x)

= 3 (xyz-x^2y-xz^2+x^2 z-y^2 z+xy^2+yz^2 -xy^2)

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

similarly x-2y+2y-3z+3z-x = 0

hence, (x-2y)^3+(2y-3z)^3+(3z-x)^3

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