Prove that $6 | (a+b+c)$ if and only if $6 | (a^3 + b^3 +c^3).$ I have tried the question but not sure if my solution is correct or not...
My try..
\begin{align}a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)\end{align}
So , if
\begin{align}6 |(a^3 + b^3 + c^3)\end{align}
Then,
\begin{align}6 | [(a+b+c)^3 - 3(a+b)(b+c)(c+a)]\end{align}
So, also
\begin{align}6 | (a+b+c)^3\end{align}
\begin{align}6 | (a+b+c)(a+b+c)(a+b+c)\end{align}
So,
\begin{align}6|(a+b+c)\end{align}
Is my solution correct ?

The reason for your incorrect solution has been pointed out by Iris in the comments.
\begin{align}
&(a^3+b^3+c^3) - (a+b+c)  = (a^3 - a) + (b^3-b) + (c^3-c) \\ &\quad \quad \ = (a-1)a(a+1) + (b-1)b(b+1) + (c-1)c(c+1)
\end{align}
The set of terms in $(a-1)a(a+1)$ contains at least one multiple of $2$ and one multiple of $3$ by the pigeonhole principle. By coprimality of $2,3$, each such number is a multiple of $6$. So $(a^3+b^3+c^3) - (a+b+c)$ is a multiple of $6$ : if one is divisible by $6$, so is the other.
