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 $k^3n−kn^3 =(k^3-k)(n-n^3)+(kn)^3+kn$
 A: $k^{3}n-kn^3 = kn(k+n)(k-n)$ 
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 $6$.
A: Here;'s a proof by induction on $n$
$$k^3-k$$ is divisible by 6 for base case $n=1$ (as you have proved)
Suppose $$k^3n-kn^3$$ is divisible by 6
Now $$k^3(n+1)-k(n+1)^3=(k^3n-kn^3)+(k^3-k)-(3kn(n+1))$$
Each of these terms are divisible by $6$
which completes induction hypothesis.
NOTE:$$(3kn(n+1)) $$
is divisible by $6$ because this expression is divisible by $3$ and there is $n(n+1)$ and two consecutive naturals are always divisible by $2$ (This fact can also be proved by induction). So, it is divisible by $6$.
