use mathematical induction to show that $n^3 + 5n$ is divisible by $3$ for all $n\ge1$ What I have so far
Base: $n^3 + 5n$
Let $n=1$
$$
1^3 + 5(1) = 6
$$
$6$ is divisible by $3$
Induction step: $(k+1)^3 + 5(k+1)$
$(k^3 + 3k^2 + 8k + 6)$ is divisible by $3$
I kind of get lost after this point. For starters, how do I prove that this isn't applicable for any number less than $1$? Also, where do I go after this? Thank you!

HINT
$$
\begin{split}
k^3 + 3k^2+8k+6
 &= 3(k^2+3k+2) + k^3-k \\
 &= 3(k^2+3k+2) + k(k^2-1) \\
 &= 3(k^2+3k+2) + k(k-1)(k+1)
\end{split}
$$
