n³+2n

If n=1 then 1³+2(1) = 1+2 = 3 which is divisible by 3.

Let us assume that n=k is some integer (perhaps 1) such that
k³+2k is divisible by n.  That is, there is some positive 
integer p such that k³+2k = 3p 
[In the case where n=1, then p=1]

We want to show that the expression n³+2n with k+1 substituted
for n also gives a multiple of 3.

We examine the case where n=k+1 and multiply it all the way out:

(k+1)³+2(k+1) = k³+6k²+11k+6.

We notice that this differs from k³+2k by 6k²+9k+6.

So we add 6k²+9k+6 to both side of

k³+2k = 3p

and get

k³+2k+6k²+9k+6 = 3p+6k²+9k+6 = 3(p+2k²+3k+2)

So (k+1)³+2(k+1) = 3(p+2k²+3k+2)

which is a multiple of 3, so the theorem is proved.

Edwin