How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3. 
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 11n + 9$
$n(n^2 + 8) + 3n^2 + 3n + 9$
How can I show that $3n^2 + 3n + 9$ is a multiple of 3?

If $n\equiv 0\pmod 3$ Ok. If $n\equiv 1\pmod 3$, we have
\begin{equation}
n^{2} + 8 \equiv 1^{2} + 2\equiv 0\pmod 3.
\end{equation}
If $n \equiv 2\pmod 3$ we have
\begin{equation}
n^{2} + 8 \equiv 2^{2} + 2\equiv 6 \equiv 0\pmod 3.
\end{equation}
