<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] Let $\wp(n) = (n-1)^3  + n^3 + (n+1)^3$. Observe that $\wp(1) = 9$, and so is divisible by $9$. To complete this inductive proof, we want to prove that $\wp(n+1) - \wp(n)$ is divisible by $9$.
$\wp(n+1) - \wp(n) = n^3 + (n+1)^3 + (n+2)^3 - (n-1)^3 - n^3 - (n+1)^3 $
$= (n+2)^3 - (n-1)^3 = 9n^2 + 9n + 9 = 9(n^2 + n +1)$
and so is also divisible by $9$. We have thus shown that $\wp(1) = 0 \mod 9$, and that $\wp(n+1) - \wp(n) = 0 \mod 9$, and so $\wp(2) = \wp(1) + \wp(2) - \wp(1) = 0 \mod 9$, and inductively proceed from there...
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