<s>[INST] If n+1 is a cube show that 504 divides n(n+1)(n+2) If $n+1$ is a cube show that 504 divides $n(n+1)(n+2)..$ 
I try that : $n+1=a^3$ , then $n=a^3-1$ so $n(n+1)(n+2) = (a^3-1)a^3(a^3+1) = (a^6-1)a^3$ and I try prove $(a^6-1)a^3$ divides by 504 with induction but it is not easy (or short). I just find anybody who solve this problem nice and short. And it is so important for me.. Please help me ...
 [/INST] You need to show that $(a^3-1)a^3(a^3+1)$ is divisible by $9$, $8$, and $7$ (since $504=9\cdot8\cdot7$).  Your idea of rewriting the product at $(a^6-1)a^3$ is a good one, since that takes care of $7$, using Fermat's little theorem.
For the other two, you can factor the product at $k^3(k-1)(k+1)(k^2+k+1)(k^2-k+1)$.  For divisibility by $8$, you either have $k$ is even, which gives $8\mid k^3$, or $k-1$ and $k+1$ are both even, in which case one of them is divisible by $4$ and the other by $2$.  Likewise, you either have $27\mid k^3$, or you have one of $k-1$ and $k+1$ divisible by $3$ and one of $k^2+k+1$ and $k^2-k+1$ divisible by $3$.
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