Answer
$g^{-1}(x)=\sqrt[3]{4^x-2}$
Work Step by Step
The inverse of $g$ can be found by solving the following equation for $x$:
$$y=\log_4(x^3+2)$$
$$y=\frac{\ln(x^3+2)}{\ln 4}$$
$$y\ln 4=\ln(x^3+2)$$
$$e^{y\ln 4}=x^3+2$$
$$e^{y\ln 4}-2=x^3$$
$$\sqrt[3]{e^{y\ln 4}-2}=x$$
$$\sqrt[3]{(e^{\ln 4})^y-2}=x$$
$$\sqrt[3]{4^y-2}=x$$
So the inverse of $g$ is:
$$g^{-1}(x)=\sqrt[3]{4^x-2}$$
