<s>[INST] Is this proof by induction correct? Prove by induction that for all $n\in\mathbb N$, $3\mid n^3+3n^2+2n$.
$$P(1) = (1)^3+3(1)^2+2(1) = 6$$
Which is clearly divisble by $3$. Therefore, $P(1)$ is true.
Assume $P(1),\ldots,P(n)$ and show $P(n+1)$ is true.
$$(n+1)^3+ 3(n+1)^2+ 2(n+1) = n^3+6n^2+11n+6$$
This can be rewritten: $(n^3+3n^2+2n)+3(n^2+3n+2)$.
First term is divisible by $3$ because $P(n)$ is true.
The second term is divisible by $3$ because it is a multiple of $3$.
Thus, their sum is divisible by $3$.
Is there anything I am missing?
 [/INST] Your proof looks solid, but I might suggest an easier proof:
For all $n \in \mathbb{N}$, we have $n \equiv 0, 1, 2 \pmod{3}$.  If $n \equiv 0 \pmod{3}$, we are done.
If $n \equiv 1 \pmod{3}$, then we have $n^3 + 3n^2 + 2n \equiv (1+2) \equiv 0 \pmod{3}$.
If $n \equiv 2 \pmod{3}$, then we have $n^3 + 3n^2 + 2n \equiv (2 + 1) \equiv 0 \pmod{3}$.
We conclude that $3|(n^3 + 3n^2 + 2n)$ for all natural numbers $n$.
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