If f : R → R is defined by

Given: $f(x)=8 x^{3}$

$f(x)=8 x^{3}$

$\Rightarrow y=8 x^{3}$

$\Rightarrow x^{3}=\frac{y}{8}$

$\Rightarrow x=\left(\frac{y}{8}\right)^{\frac{1}{3}}$

Thus, $f^{-1}(x)=\left(\frac{x}{8}\right)^{\frac{1}{3}}$

$f^{-1}(8)=\left(\frac{8}{8}\right)^{\frac{1}{3}}$

$=1^{\frac{1}{3}}$                 $(\because f: R \rightarrow R)$

Hence, if $f: R \rightarrow R$ is defined by $f(x)=8 x^{3}$ then $f^{-1}(8)=1$.

$=1$