Question:
If $f: C \rightarrow C$ is defined by $f(x)=8 x^{3}$, then $f^{-1}(8)=$ . _________.
Solution:
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, $f^{-1}(x)=\left(\frac{x}{8}\right)^{\frac{1}{3}}$
$f^{-1}(8)=\left(\frac{8}{8}\right)^{\frac{1}{3}}$
$=1^{\frac{1}{3}}$
$=1, \omega, \omega^{2} \quad(\because f: C \rightarrow C)$
where, $\omega$ is the cube root of unity.
Hence, if $f: C \rightarrow C$ is defined by $f(x)=8 x^{3}$, then $f^{-1}(8)=1, \omega, \omega^{2} .$
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