N is a natural number such that when N³ divided by 9 ,it leaves remainder a.It can be concluded that

a)a is a perfect square b) a is a perfect cube c) both of these d ) none of these

a is a perfect cube.
by the Euclid's Lemma any number can be represented as either of 3k , 3k+1 or 3k+2 form
case 1: let N = 3k
$N^3=(3k{)}^3=9*3k^3$ which is divisible by 9. thus remainder a = 0
case 2: let N = 3k+1
$$\begin{gathered} N^3=(3k+1{)}^3=(3k{)}^3+1^3+3*(3k{)}^2*1+3*(3k)*1^2 \\ =27k^3+27k^2+9k+1 \\ =9(3k^3+3k^2+k)+1 \end{gathered}$$ 
thus when $N^3$ is divided by 9, remainder a = 1, which is a perfect cube.
case 3: let N = 3k+2
$$\begin{gathered} N^3=(3k+2{)}^3=(3k{)}^3+2^3+3*(3k{)}^2*2+3*3k*2^2 \\ =27k^3+8+54k^2+36k \\ =9(3k^3+6k^2+4k)+8 \end{gathered}$$
thus when $N^3$ is divided by 9, remainder a = 8 which is a perfect cube.
hope this helps you.