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 ?

Since $(n-1)^3 + n^3 + (n+1)^3 = 3n(n^2+2)$ we know that it is divisible by $3$. If we show that $n(n^2+2)$ is also divisible by $3$ then $(n-1)^3 + n^3 + (n+1)^3$ is divisible by $3^2 = 9$. There are three cases to check:


*

*$0(0^2+2) \equiv 0 \cdot 2 \equiv 0 \pmod 3$

*$1(1^2+2) \equiv 1 \cdot 0 \equiv 0 \pmod 3$

*$2(2^2+2) \equiv 2 \cdot 0 \equiv 0 \pmod 3$

