I am trying to use induction. Have I applied it correctly / rigorously enough?
Prove that the sum of three consecutive cubes are divisible by $9$.
Base case: Let $n=0$. Then $0^3 + 1^3 + 2^3 \equiv 9 \equiv 0 \bmod 9$, and so this is true.
Inductive phase: Suppose $n^3 + (n+1)^3 + (n+2)^3 \equiv 0 \bmod 9$. We want to prove that $(n+1)^3 + (n+2)^3 + (n+3)^3 \equiv 0 \bmod 9$.
Subtracting one from the other, we get $(n+3)^3 - n^3 \equiv 0 \bmod 9$.
This expands to $9 n^2+27 n+27 \equiv 0 \bmod 9$.
Divide both sides and modulus by $9$:
$n^2+3n+3 \equiv 0 \bmod 1$, which is true because $1$ divides all positive integers.
Therefore this completes the proof.
Did I miss anything? Is my proof complete? What could I do better?