I'd use the induction hypothesis to prove it.
Basically the induction hypothesis shows that if something works for n then it will work for n + 1, which then shows it will work for all values of n.
That should explain it in more detail.
First of all you need a basis, to do this show that your hypothesis (that its divisible by 3 for all values of n) works for a small value of n - you can use either n = 0 or n = 1.
Then you substitute in n+1 for n in your first expression.
So in full your proof should look a bit like:
For any natural number n, n3
+ 2n is divisible by 3.
Basis: If n = 0, n3
+ 2n = 03
+ 2×0 = 0.
0 mod 3 = 0, therefore it is divisible by 3.
Induction: Assume that for an arbitrary natural number n, n3 + 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 n3
+ 2n and use the induction hypothesis.
( n + 1 )3
+ 2( n + 1 ) = ( n3
+ 3n + 1 ) + ( 2n + 2 )
= ( n3
+ 2n ) + ( 3n2
+ 3n + 3 )
= ( n3
+ 2n ) + 3( n2
+ n + 1 )
+ 2n is divisible by 3 by the induction hypothesis.
I'm not great with proofs so prehaps someone else could just check that.