**Given**

n^{3} + 2n is divisible by 3 ∀ n ∈ N

**Proof**

We will prove the given statement by induction

**STEP 1**

n = 1

n^{3} + 2n = 1^{3} + 2*1 = 3

3 is divisible by 3. Therefore, the statement is true for n = 1

**STEP 2**

Let the given statement be true for n = k

k^{3} + 2k = 3x

Now, we need to prove that if the statement is true for n = k then it is also true for n = k + 1

(k + 1)^{3} + 2(k + 1) = k^{3} + 1 + 3k^{2} + 3k + 2k + 2

= (k^{3} + 2k) + 1 + 3k^{2} + 3k + 2

= 3x + 3k^{2} + 3k + 3

= 3 * ( x + k^{2} + k + 1 )

Thus, we can say that (k + 1)^{3} + 2(k + 1) is divisible by 3 if k^{3} + 2k is divisible by 3.

Therefore, we can say that if the given statement is true for n = k, then it is also true for n = k + 1. Hence, by the principle of mathematical induction, the given statement is true ∀ n ∈ N.