How to prove that $k^3+3k^2+2k$ is always divisible by $3$? How can I prove that the following polynomial expression is divisible by 3 for all integers $k$?
$$k^3 + 3k^2 + 2k$$

Just to be different.  If $k \in \mathbb Z$ then $k = 3m + i$ where $i = $ either $0, 1,$ or $-1$ and $m \in \mathbb Z$.
So
\begin{align*}
k^3 + 3k^2 + 2k
&=(3m + i)^3 + 2(3m + i)+ 3k^2 \\
&= 3^3m^3 + 3 \cdot 3^2m^2 \cdot i + 3 \cdot 3m \cdot i^2 + i^3 + 2 \cdot 3m + 2i + 3k^2 \\
&= i^3 + 2i + 3\big[3^2m^3 + 3^2m^2i + 3m \cdot i^2 + k^2\big]
\end{align*}
Now $i^3 = i$ for $i = 0,1,-1$ so $i^3 + 2i = 3i$.
So
$$
k^3 + 3k^2 + 2k = 3\big[i + 3^2m^3 + 3^2m^2i + 3m \cdot i^2 + k^2\big].
$$
...
But seriously, factoring is the better way to do it.
