Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!

Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
  This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
  $2 \times 0 = 0.$ So it is divisible by $3.$
Induction: Assume that for an arbitrary natural number $n$,
  $n^3+ 2n$ is divisible by $3.$ 
Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
  the induction hypothesis. Got it
$$( n + 1 )^3+ 2( n + 1 ) = (  n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
$$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
 and regrouping}\}$$
  $$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
  by the induction hypothesis. What?

Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$

We can take three cases (it's worth reading up on this by the way, the idea is called 'modular arithmetic)
$n≡0$ mod $3$
$n≡1$ mod $3$
$n≡2$ mod $3$
In the first case we are immmediately done because $n^3+2n=n(n^2+2)$ and since n factors out, by assumption we are done. 
In the second and thirdcase we will substitute $'n'$ by it's respective residue in the equation and hope that the outcome will equal $0 mod 3$
$n≡1$ mod $3$ gives us $1^3$ mod $3$ $+ 1*2$ mod $3$ $≡ 1+2$ mod $3$ $≡ 0$ mod $3$
$n≡2$ mod $3$ gives us $2^3$ mod $3$ $+ 2*2$ mod $3$ $≡ 8+4$ mod $3$ $≡ 12$ mod $3$ $≡ 0$ mod $3$
