$24\mid n(n^{2}-1)(3n+2)$ 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.

To prove $8|n(n^2-1)(3n+2)$ you can simply break the problem in three cases:
Case 1 $n$ odd.
Then $n^2-1=(n-1)(n+1)$ is the product of two consecutive even numbers. Thus one must be divisible by $4$ and the other by $2$.
Case 2 $n$ multiple of 4.
Then $4|n$ and $2|3n+2$.
Case 3 $n$ even but not divisible by 4.
Then $2|n$ and $n-2$ is divisible by $4$. Hence 
$$4|3n+2=4n-(n-2) \,.$$
