<s>[INST] $24\mid n(n^{2}-1)(3n+2)$ 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$$8\mid n(n^{2}-1)(3n+2)?$$$$$$
Would not, ever succeeded without the help of everyone that this will post tips, ideas, etc., etc.. Thank you.
 [/INST] We know from here or here, the product $n$  consecutive integers is divisible by $n!$ for integer  $n>0$
As $24=4!$and we already have $n(n^2-1)=(n-1)n(n+1)$ the product of $3$ consecutive integers
So, if we can arrange the next or previous term as multiplier, we shall have the product of $4$ consecutive integers, hence divisible by $4!=24$
If we consider the previous term $(n-2),$ 
$3n+2=3(n-2)+4$
$$\implies n(n^2-1)(3n+2)=n(n^2-1)\{3(n-2)+4\}$$
$$=3\underbrace{(n+1)n(n-1)(n-2)}_{\text{ the product of } 4 \text{ consecutive integers }} +4 \underbrace{(n+1)n(n-1)}_{\text{ the product of } 3\text{ consecutive integers }}$$
If we consider the next term $(n+2),$ 
$3n+2=3(n+2)-4$
$$\implies n(n^2-1)(3n+2)=n(n^2-1)\{3(n+2)-4\}$$
$$=3\underbrace{(n+2)(n+1)n(n-1)}_{\text{ the product of } 4 \text{ consecutive integers }}-4\underbrace{(n+1)n(n-1)}_{\text{ the product of } 3\text{ consecutive integers }}$$
</s>