Question

show that (a-b)3+(b-c)3+(c-a)3=3(a-b)(b-c)(c-a)

Answer 1

LHS=(a-b)3+(b-c)3+(c-a)3
=3a-3b+3b-3c+3c-3a
=0
RHS=3 (a-b)(b-c)(c-a)
=3 ((ab-ac-b^2+bc)(c-a))
=3 (abc-a^2b-ac^2+a^2c-cb^2+ab^2+bc^2-abc)
=3 (0)
=0
LHS=RHS