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?

Using induction it is obvious that the statement is true for $n=1$. 
Now suppose that it is true for $n=k$,
then we have $k(k^2+8)=3m$, where m is an integer.
Considering the case where $n=k+1$, we have;
$$(k+1)[(k+1)^2+8]=k(k^2+2k+9)+k^2+2k+9$$
$$=k(k^2+8+2k+1)+k^2+2k+9=k(k^2+8)+k(2k+1)+k^2+2k+9$$
$$=k(k^2+8)+3k^2+3k+9=3m+3k^2+3k+9=3(m+k^2+k+3)$$
