<s>[INST] 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 ?
 [/INST] I'm not sure how you got

Then,
6 | [(a+b+c)^3 - 3(a+b)(b+c)(c+a)]
So, also
6 | (a+b+c)^3
6 | (a+b+c)(a+b+c)(a+b+c)

As Iris's question comment says, because $6$ divides the first expression of $(a+b+c)^3 - 3(a+b)(b+c)(c+a)$ doesn't mean it divides the first term, i.e., $(a+b+c)^3$ and then a different expression of $(a+b+c)(a+b+c)(a+b+c)$.
Instead, here is a simpler way. Note that odd integers cubed are odd & even integers cubed are even, so for all integers $n$ you have
$$n^3 \equiv n \pmod 2 \tag{1}\label{eq1A}$$
Also, as $3$ is prime, then by Fermat's little theorem, for all integers $n$,
$$n^3 \equiv n \pmod 3 \tag{2}\label{eq2A}$$
Since $2$ and $3$ are relatively prime, you can put these together to get
$$n^3 \equiv n \pmod 6 \tag{3}\label{eq3A}$$
Thus, you get
$$a + b + c \equiv a^3 + b^3 + c^3 \pmod 6 \tag{4}\label{eq4A}$$
As such,
$$a + b + c \equiv 0 \pmod 6 \iff a^3 + b^3 + c^3 \equiv 0 \pmod 6 \tag{5}\label{eq5A}$$
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