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!

Just note that
$$
f(n+1)-f(n)=(n+1)^3 + 5(n+1)-(n^3 + 5n)=3 (n^2 + n + 2)
$$
I suspect this is the intended solution.
Actually, no induction is needed:
$$
n^3 + 5n = 6 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}
$$
and so is always a multiple of $6$.
