solve a math question.?

Question

suppose:a+b+c=0
prove:a^3+b^3+c^3=3abc

Answer 1

Given a+b+c=0
a+b=-c
Cubing both sides, we get
(a+b)^3=(-c)^3
a^3+3a^2 b+3ab^2+b^3=-c^3
a^3+b^3+c^3=-3a^2 b-3ab^2=-3ab(a+b)=-3ab(-c)=3abc
Therefore a^3+b^3+c^3=3abc

Answer 2

Factorize a^3+b^3+c^3- 3abc..
you will obtain:
a^3+b^3+c^3- 3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

now from this when a+b+c =0.
you will have a^3+b^3+c^3 = 3abc

I hope you are able to factorize it, in order to prove it that way.

Answer 3

I agree with K N Swamy