Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$ 
Question-Prove that $6|a+b+c $ if and only if $6|a^3 +b^3+c^3$

I was playing around with the formulae
$$(a+b+c)^3=a^3 +b^3+c^3+3(a+b)(b+c)(c+a)$$ and,
$$a^3 +b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
and noted that if $(a+b+c)\equiv0$(mod 6)$\implies a^3 +b^3+c^3\equiv3abc$(mod 6). Now I am not sure how to show $3abc\equiv 0$(mod6), and even doing that, we have only half of the proof because then we need to prove that the converse is also true.

Among $a$, $b$, $c$ there are at least two even or two odd numbers. In any case $(a+b)(b+c)(c+a)$ is even.
