Check the injectivity and surjectivity of the following function $f : N \rightarrow N$ given by $f(x) = x^{3}$.

(N/A) Given function is $f : N \rightarrow N$ defined by $f(x) = x^{3}$.
For injectivity:
Let $x, y \in N$ such that $f(x) = f(y)$.
Then $x^{3} = y^{3}$.
Since $x, y$ are natural numbers,taking the cube root on both sides gives $x = y$.
Therefore,$f$ is injective.
For surjectivity:
$A$ function is surjective if for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$.
Consider $y = 2 \in N$.
We need $x^{3} = 2$,which implies $x = \sqrt[3]{2}$.
Since $\sqrt[3]{2}$ is not a natural number $(\sqrt[3]{2} \notin N)$,there is no $x \in N$ such that $f(x) = 2$.
Therefore,$f$ is not surjective.
Conclusion: The function $f$ is injective but not surjective.