TITLE: Proof that $n^3 + 3n^2 + 2n$ is a multiple of $3$.
QUESTION [12 upvotes]: I'm struggling with this problem:

For any natural number $n$, prove that $n^3 + 3n^2 + 2n$ is a multiple of $3$.

That $n^3 + 3n^2 + 2n$ is a multiple of $3$ means that: $n^3 + 3n^2 + 2n = 3 \times k$ where $k \in \mathbb N$
So I tried to find a the number $k$.
The best result I found was: $n^3 + 3n^2 + 2n = n(n+1)(n+2)$ 
But I'm lagging at the last step, to prove that it is a multiple of $3$.
(However, I got the intuition, If you see the multiples of $3$: $\{0, 3, 6, 9,\dotsc\}$ there is a difference of $3$ between them.
So $n(n+1)(n+2)$ incorporates that difference. This means that if you take a number $n$ then $n$ or $n+1$ or $n+2$ could be a multiple of $3$ and so their multiplication is a multiple of $3$) But I couldn't extend that idea into a consistent mathematical proof.
Also this problem doesn't help either:
Proof that $n^3+2n$ is divisible by $3$

REPLY [2 votes]: In response to your inductive proof link to another example, I thought you might find a similar proof useful to prove that $n^3 + 3n^2 + 2n = n(n+1)(n+2)$ is a multiple of $3$ (I will say divisible by 3, you can modify it yourself if you care).
$\textbf{Proof:}$ (by induction on $n$):
For the basis $n=1$, the result is trivial.
Assume the divisibility holds for $n=k$, that is,
$$3 \, | \, k(k+1)(k+2).$$
Let $n=k+1$. Then
$$(k+1)(k+2)(k+3)=k(k+1)(k+2) + 3(k+1)(k+2).$$
Now clearly 
$$3 \, | \, 3(k+1)(k+2),$$ 
and from our inductive hypothesis, we have that 
$$3 \, | \, k(k+1)(k+2).$$
Hence
 $$3 \, | \, (k(k+1)(k+2) + 3(k+1)(k+2)).$$ 
Therefore $3 \, | \, n(n+1)(n+2)\, \forall \, n \in \mathbb{N}$.