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?
You've already solved the case $n=1$, so I'll not repeat that there.

Assuming as the induction hypothesis that $n$ has the property that $3|n(n^2+8)$, we can rewrite $(n+1)((n+1)^2+8)$ to obtain
$$
\begin{split}
(n+1)((n+1)^2+8) &amp; = (n+1)(n^2+2n+1+8)\\
&amp; =n(n^2+8)+n(2n+1)+((n+1)^2+8)\\
&amp; = n(n^2+8)+3n^2+3n+9
\end{split}
$$
In the latter, all the terms are divisible by $3$, hence it follows that $(n+1)((n+1)^2+8)$ is also divisible by $3$. This finishes the induction proof, so we may conclude that $n(n^2+8)$ is divisible by $3$ for all $n\geq 1$.
Quote: How can I show that $3n^2 + 3n + 9$ is a multiple of $3$? End of quote

I'm surprised at the complexity of some of the answers.

$$
3n^2 + 3n + 9 = 3(n^2 + n + 3) = (3\cdot\text{something}).
$$

A "multiple of $3$" is anything that is $3$ times something (where "something" means of course an integer).