Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ 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 ?

From your computations, it's enough to show that $n^3+2n=n(n^2+2)\equiv 0\pmod{3}$. If $n\equiv 0\pmod{3}$, you are done, else $n\equiv 1,2\pmod{3}$, in which case $n^2+2\equiv 0\pmod{3}$. 
Reducing it to modular arithmetic may save you a lot of messy multiplication.
