Proving $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ I'm trying to prove by MI. I have already distributed n+1, but now I'm stuck on how I can show 9 divides the RHS since $42n$ and $3n^3$ does not divide evenly. 
$$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36$$

your term is equivalent to
$3(n+1)(n^2+2n+3)$
if $n\equiv 0\mod 3$ then $n^2+2n+3\equiv 0 \mod 3$
if $n\equiv 1 \mod 3$, then $n^2+2n\equiv 0 \mod 3$
if $n\equiv 2 \mod 3$ then $n+1 \equiv 0 \mod 3$
