<s>[INST] Prove that$ k^3n-kn^3$ is divisible by $6 $ for all n∈N. Hello I have problem with solution of task.
Prove that $k^{3}n-kn^3$ is divisible by $6$ for all $n∈N$, $k∈N$ .
Help me, please.
I know, when $n^3-n$ is divisible by 6. 
$n^3-n= (n-1)(n)(n+1)$ and is divisible.
I having similar idea $(kn)^3-(kn)= (kn-1)(kn)(kn+1)$
and my 2 idea is $k^3n−kn^3 =(k^3-k)(n-n^3)+(kn)^3+kn$
 [/INST] $k^{3}n-kn^3 = kn(k+n)(k-n)$ 
If this is divisible by both $2$ and $3$, then it is divisible by $6$.
Assuming $k$ is an integer, if either $k$ or $n$ is even, then the expression is even, and if they are both odd, then $k+n$ is even, so the expression is still even.
If either $k$ or $n$ are divisible by $3$, then the expression is divisible by $3$.
If $k$ and $n$ are either both $\equiv 1\pmod 3$ or both $\equiv 2\pmod 3$, then $k-n$ will be divisible by $3$.
If one of the two variables is $\equiv 1\pmod 3$ and the other is $\equiv 2\pmod 3$ then $k+n$ will be divisible by $3$.
So in all scenarios, the expression is divisible both by $2$ and $3$ and therefore divisible by $6$.
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