Proof that $3\mid n^3 − 4n$ Prove that $n^3 − 4n$ is divisible by $3$ for every positive integer $n$.
I am not sure how to start this problem. Any help would be appreciated
 A: Well, one can start off by factoring $n^3 - 4n$:
$n^3 - 4n = n(n^2 - 4)$
$= n(n + 2)(n - 2); \tag{1}$
next we note that every $n \in \Bbb Z$ is of one of the three forms
$n = 3q, \tag{2}$
$n = 3q + 1, \tag{3}$
or
$n = 3q + 2; \tag{4}$
this follows from the basic properties of Euclidean division applied to $n$ with $3$ as the divisor; the only possible remainders $r$ be $0$, $1$, and $2$, since such integers $r$ must satisfy $0 \le r \le 2$.  In case (2), clearly $3 \mid n$; in case (3), we have
$n + 2 = 3q + 1 + 2$
$= 3q + 3 = 3(q + 1), \tag{5}$
showing that $3 \mid n + 2$; finally, case (4) yields
$n - 2$
$= 3q + 2 - 2 = 3q, \tag{6}$
whence $3 \mid n - 2$.  For evey possible $n$, $3$ divides one of the factors of $n^3 - 4n$ occurring in (1); thus we have $3 \mid n^3 - 4n$ for all $n \in \Bbb Z$.  QED.
A: $n^3-n-3n = n(n^2-1)-3n = n(n-1)(n+1)-3n$. As, $3|3n$ and $3|(n-1)$ or $3|n$ or $3|(n+1) \implies 3|n^3-4n$
A: Note that
$$
\begin{align}
n^3-4n
&=6\binom{n}{3}+6\binom{n}{2}-3\binom{n}{1}\\
&=3\left[2\binom{n}{3}+2\binom{n}{2}-\binom{n}{1}\right]
\end{align}
$$
