f:Z→Z is given by f(x) = x3 is-

f:Z→Z   is given by f(x) = x³  

It is seen that for x, y ∈ Z , f(x) = f(y)  ⇒ x³ = y³ ⇒ x =y

So f is injective.

Now, 2 ∈ Z, But there does not exist any element x in Z such that f(x) = x³ = 2

so f is not surjective.

Hence, function f is injective but not surjective.