Answer
$f^{-1}(x)=\sqrt[3]{\dfrac{x-8}{3}}$
Work Step by Step
$f(x)=3x^{3}+8$
Rewrite this expression as $y=3x^{3}+8$ and solve for $x$:
$y=3x^{3}+8$
$3x^{3}+8=y$
Take $8$ to the right side:
$3x^{3}=y-8$
Take the $3$ to divide the right side:
$x^{3}=\dfrac{y-8}{3}$
Take the cubic root of both sides:
$\sqrt[3]{x^{3}}=\sqrt[3]{\dfrac{y-8}{3}}$
$x=\sqrt[3]{\dfrac{y-8}{3}}$
Interchange $x$ and $y$:
$y=\sqrt[3]{\dfrac{x-8}{3}}$
The inverse of the initial function is $f^{-1}(x)=\sqrt[3]{\dfrac{x-8}{3}}$
