<s>[INST] 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 ?
 [/INST] To say that $3n^3+6n$ is divisible by $9$ is equivalent to saying that $n^3+2n=n(n^2+2)$ is divisible by $3$. Now split to three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$ (explain why this is true.) Now you have to check just 3 cases.
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