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 $k$?
$$k^3 + 3k^2 + 2k$$

Without noticing that the expression is factorisable into 3 consecutive numbers:
Define $a = k \;\text{mod}\;3$
$$(k^3 + 3k^2 + 2k) \;\text{mod}\;3 \\
= (k^3\;\text{mod}\;3 + 3k^2\;\text{mod}\;3 + 2k\;\text{mod}\;3) \;\text{mod}\;3 \\
= (a^3\;\text{mod}\;3 + 3a^2\;\text{mod}\;3 + 2a\;\text{mod}\;3) \;\text{mod}\;3$$
$a^3 \;\text{mod}\;3$:


*

*If $a=1$, $a^3 = 1$. 

*If $a=2$, $a^3\;\text{mod}\;3=8\;\text{mod}\;3=2$. 


So $a^3 \;\text{mod}\;3 = a$.
$3a^2\;\text{mod}\;3 = 0$
$2a\;\text{mod}\;3$:


*

*If $a=1$, $2a\;\text{mod}\;3 = 2$.

*If $a=2$, $2a\;\text{mod}\;3 = 4\;\text{mod}\;3 = 1$. 


So $2a \;\text{mod}\;3 = -a\;\text{mod}\;3$.
$$(a + 0 + -a)\;\text{mod}\;3 = 0$$.
