$n^{3} + 2n$ is divisible by $3$. Is this induction proof correct? Question: Prove by means of the principle of induction that for every $n ∈ N$ the
number $n^{3} + 2n$ is divisible by $3$.
Proof
Denote "$n^{3} + 2n$ is divisible by 3" by $P(n)$. Check $P(n)$ for an arbitrary $n$, for instance $n=1$. $1^{3}+2*1=3*1$ and thus divisible by three, therefore $P(1)$ holds.
Induction step: Assume $P(n)$ is is true, let $n ∈ N$. Then $(n+1)^{3}+2(n+1) =(n+1)(n^{2}+2n+1)+2(n+1)=(n+1)(n^{2}+2n+3)=n^{3}+2n^{2}+3n+n^{2}+2n+3=n^{3}+2n+3(n^{2}+n+1).$
We assumed $P(n)$ holds, thus the $n^{3}+2n$ part of the induction holds. We can obviously see that $3(n^{2}+n+1)$ is divisible by 3 and that concludes the proof.

I suggest a slight rewording.
Proof
Denote the statement

$n^{3} + 2n$ is divisible by 3

by $P(n)$. We check $P(n)$ for an  $n=1$: $1^{3}+2*1=3*1$. Thus $P(1)$ holds.
Induction step: Assume $P(n)$ is is true, let $n ∈ N$. Then $(n+1)^{3}+2(n+1) =(n+1)(n^{2}+2n+1)+2(n+1)=(n+1)(n^{2}+2n+3)=n^{3}+2n^{2}+3n+n^{2}+2n+3=n^{3}+2n+3(n^{2}+n+1).$
We assumed $P(n)$ holds, thus $n^{3}+2n$ is divisible by $3$. We can obviously see that $3(n^{2}+n+1)$ is divisible by 3 and that concludes the proof.
