If $f:R\to R$ defined by $f\left(x\right)=x^3,$ then $f$ is

$f:R\to R,f\left(x\right)=x^3$

To prove the $\mathrm{f}$ is bijective we must prove that $\mathrm{f}$ is one-to-one and onto Proof $\mathrm{f}$ is one-to-one

Let $x_1,x_2\in R$

$f\left(x_1\right)=f\left(x_2\right)$

$x_1^3=x_2^3\Rightarrow x_1=x_2$

Since the cube root is a function we can take the cube root of both sides We have now shown that $\mathrm{f}$ is one-to-one

$\mathrm{f}$ is onto

Let $y\in R$

$x^3=y$

$x=\sqrt[3]{y}$

$x$is clearly a real number since cube root is closed over the reals and $f\left(x\right)={\left(\sqrt[3]{y}\right)}^3=y$ thus have found the required pre-image for $\mathrm{f}$ and $\mathrm{f}$ is onto by definition.