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.
 A: Since $x^3\equiv_3 x$ for each integer $x$ we have  $$3|a+b+c \iff a+b\equiv_3 -c$$ $$\iff (a+b)^3\equiv_3 (-c)^3$$ 
$$\iff a^3+3ab(a+b)+b^3\equiv_3 -c^3$$ 
$$\iff a^3+b^3+c^3\equiv_3  0 $$ $$\iff 3\mid a^3+b^3+c^3$$
A: Hint:
$$a^3-a=(a-1)a(a+1)$$ is divisible by $6$ being the product of three consecutive integers
$$\implies a^3+b^3+c^3+\cdots\equiv a+b+c+\cdots\pmod6$$ 
A: Now, use that $a+b+c$ is an even number.
Can you end it now?
A: 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.
A: a^3+b^3+c^3-a-b-c = (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1)

(a-1)a is divisible by 2.
(a-1)a(a+1) is divisible by 3.
Hence, (a-1)a(a+1) is divisible by 6.
Similarly, (b-1)b(b+1) and (c-1)c(c+1) are also divisible by 6.
Since a^3+b^3+c^3 is divisible by 6 and (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1) is divisible by 6,
a+b+c is also divisible by 6.
For the converse,
a+b+c is divisible by 6, (a-1)a(a+1)+(b-1)b(b+1)+(c-1)c(c+1) is also divisible by 6,
hence a^3+b^3+c^3 is also divisible by 6.
I hope that this is the correct answer.
