**Given**

n^{3} + (n + 1)^{3} + (n + 2)^{3} is divisible by 9 ∀ n ∈ N

**Proof**

We will prove the given statement by induction

**STEP 1**

n = 1

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

= 1 + 8 + 27

= 36

36 is divisible by 9. Therefore, the statement is true for n = 1

**STEP 2**

Let the given statement be true for n = k

k^{3} + (k + 1)^{3} + (k+ 2)^{3} = 9x

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

= 9x – k^{3} + (k+ 3)^{3}

= 9x – k^{3} + k^{3} + 3^{3} + 3*k^{2}*3 + 3*k*3^{2}

= 9x + 27 + 9k^{2} + 27k

= 9( x + 3 + k^{2} + 3k )

9( x + 3 + k^{2} + 3k ) is divisible by 9.

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.