**Given**

n^{3}+ 2 is not divisible by 8 ∀ n ∈ N

**Proof**

For value of n, we have 2 cases

**CASE 1 : n is odd**

If n is odd then n^{3} is also odd. We know ODD + EVEN = ODD. Therefore, n^{3}+2 is odd. 8 cannot divide an odd number. We can say that n^{3}+ 2 is not divisible by 8 ∀ n ∈ O.

**CASE 2 : n is even**

If n is even, then we can write n as

n = 2^{p}x, p ∈ N

n^{3}+ 2 = (2^{p}x)^{3}+ 2

= 2^{3p} * x^{3} + 2

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

The term 2^{3p-1} * x^{3} is even. Therefore, 2^{3p-1} * x^{3} + 1 is odd. Thus, 2 ( 2^{3p-1} * x^{3} + 1 ) cannot be divisible by 8. We can say that n^{3}+ 2 is not divisible by 8 ∀ n ∈ E.

From the above two cases, we can conclude that n is not divisible by 8 ∀ n ∈ N.