Answer
$$g^{-1}(x)=\sqrt[3]{4^x-2}$$.
Work Step by Step
Given
$$ g(x)=\log _4\left(x^3+2\right) $$
Let
\begin{aligned}
y&= \log _4\left(x^3+2\right)\\
4^y&= x^3+2\\
4^y-2&=x^3\\
\end{aligned}
Hence
$$ x=\sqrt[3]{4^y-2}$$
Interchange $x$ and $$y: y=\sqrt[3]{4^x-2}$$.
It follows that
$$g^{-1}(x)=\sqrt[3]{4^x-2}$$.
