How to prove that $k^3+3k^2+2k$ is always divisible by $3$?

Rewrite $k^3 + 3k^2 + 2k = k(k+1)(k+2)$. Since exactly one of the three factors must be divisible by 3, the product must also.

HINT: Try factoring!

Another hint:

$$k^3 + 3k^2 + 2k = k(k^2 + 3k + 2) = k (k+1)(k+2) $$ Thus, you expression is the product of three consecutive integers.

To check if values $p(k)$ of the polynomial $p$ with integer coefficients is divisible by $m$ for all integer $n$, you only to check that $$p(k) \equiv 0 \pmod m, \; \forall k \in {0,\ldots,n-1}\tag{1}$$ or equivalently $$m \mid p(k), \; \forall k \in {0,\ldots,n-1}$$

So if $p(k)=k^3+3k^2+2k$ we have

• $p(0)=0$ is divisible by $3$
• $p(1)=6$ is divisible by $3$
• $p(2)=24$ is divisible by $3$

and therefore $p(k)$ is divisible by $3$ for every integer $k$.

I think using $(1)$ is less elegant than the solution that factors the polynomial $p(k)$ but it shows a statement about the infinite set of integers that can be divided in finitely many cases that can be checked by a computer program.