Simple Proof by induction: $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:


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*Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid

*Inductive step: let $P(n)$ be the statement $9$ divides $n^3 + (n+1)^3 + (n+2)^3$. Assume $P(k)$ is true, so $9$ divides $k^3 + (k+1)^3 + (k+2)^3$.


And this is where I'm stuck. I'm not sure how to demonstrate that $9$ divides $P(n)$ when $n = k+1$. If someone could step me in the right direction that would be awesome. Thanks!

$$n^3+(n+1)^3+(n+2)^3=n^3+(n+1)^3+(n+2)^3-3n(n+1)(n+2)+3n(n+1)(n+2)\equiv$$
$$\equiv n^3+(n+1)^3+(n+2)^3-3n(n+1)(n+2)=$$
$$=(n+n+1+n+2)(n^2+(n+1)^2+(n+2)^2-n(n+1)-n(n+2)-(n+1)(n+2))=$$
$$=3(n+1)\cdot3=9(n+1)\equiv0.$$
Done!
