Why is $2x^3 + x$, where $x \in \mathbb{N}$, always divisible by 3? So, do you guys have any ideas? Sorry if this might seem like dumb question, but I have asked everyone I know and we haven't got a clue.

$x \mod 3=\{0,1,2\}$ for $x\in \mathbb{N}$. 
In the other words, if you divide any natural number by $3$, the remainder will be $0$ or $1$ or $2$.


*

*$x \mod 3=0 \Rightarrow (2x^3 + x) \mod 3 = 0$

*$x \mod 3=1 \Rightarrow (2x^3 + x) \mod 3 = 0$

*$x \mod 3=2 \Rightarrow (2x^3 + x) \mod 3 = 0$

