$24\mid n(n^{2}-1)(3n+2)$ for all $n$ natural problems in the statement. "Prove that for every $ n $ natural, $24\mid n(n^2-1)(3n+2)$"
Resolution:
$$24\mid n(n^2-1)(3n+2)$$if$$3\cdot8\mid n(n^2-1)(3n+2)$$since$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)\Rightarrow3\mid n(n^{2}-1)(3n+2)$$and$$8\mid n(n^{2}-1)(3n+2)?$$$$$$
Would not, ever succeeded without the help of everyone that this will post tips, ideas, etc., etc.. Thank you.

Another approach, perhaps more elementary (but lengthier):
$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)$$
(1) Clearly, exactly one of $\,n-1\,,\,n\,,\,n+1\;$ is divisible by $\,3\,$
(2) If $\,n\,$ is odd, then $\,n-1\,,\,n+1\;$ are even with exactly one of them divisible by $\,4\,\implies (n-1)n(n+1)\,$ is divisible by $\,2\cdot 4=8\;$
(3) If $\,n\,$ is even, then also $\;3n+2\;$ is even, and exactly one of them is divisible by $\,4\,$ , since
$$n=4k+2\implies 3n+2=12k+8=4(3k+2) $$
and again $\;n(3n+2)\;$ is divisible by $\;4\cdot 2=8\;$.
The above cover all the possible cases and in each one the given expression is divisible by $\;3\;$ and by $\;8\;$ , so also by $\,3\cdot 8 =24\;$
