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 ?

$9|3n^3 + 6n$ , is equivalent to , $3|n^3 + 2n$ , is equivalent to $3|n^3 - n + 3n$ , is equivalent to ,
$3|n^3 - n$ ; now $n^3 - n = n(n-1)(n+1)$ is the product of three consecutive  integers whenever $n$ 
is an integer and since consecutive multiples of $3$ occur with a difference(gap) of $3$ , 
$n(n-1)(n+1)=n^3 - n$ must be divisible by $3$ whenever $n$ is an integer.
