Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$. Therefore $c$ should be integers, but I don't know how to do it at next step ?