First, prove the statement is true for n = 1.
For n = 1, n^3 + 2n = 1^3 + 2(1) = 1 + 2 = 3.
3 is divisible by 3, so the statement is true for n = 1.
Assume that for a positive integer k, k^3 + 2k is divisible by 3.
Then for k + 1, (k + 1)^3 + 2(k + 1) is also divisible by 3.
(k + 1)^3 + 2(k + 1) =
= k^3 + 3k^2 + 3k + 1 + 2k + 2
= k^3 + 2k + 3k^2 + 3k + 3
= (k^3 + 2k) + 3(k^2 + k + 1)
According to our assumption, k^3 + 2k is divisible by 3, and the product 3(k^2 + k + 1) must be divisible by 3 because 3 is a factor in the product. The sum of two multiples of 3 is a multiple of 3, so the expression works for k + 1.
