Show that the function f : R $\rightarrow$ R given by f(x) $= x^3$ is injective.

Let ${x_1},{x_2} \in R$ be such that,

$f({x_1}) = f({x_2}) \Rightarrow x_1^3 = x_2^3 \Rightarrow {x_1} = {x_2}$

$\therefore$ f is one$-$one. Hence, f(x) $= x^3$ is injective.