Answer
$f^{-1}(x)=\sqrt[3]{\dfrac{5-x}{4}}$
Work Step by Step
$f(x)=5-4x^{3}$
Rewrite this expression as $y=5-4x^{3}$ and sokve for $x$:
$y=5-4x^{3}$
Take $-4x^{3}$ to the left side and $y$ to the right side:
$4x^{3}=5-y$
Take the $4$ to divide the right side:
$x^{3}=\dfrac{5-y}{4}$
Take the cubic root of both sides:
$\sqrt[3]{x^{3}}=\sqrt[3]{\dfrac{5-y}{4}}$
$x=\sqrt[3]{\dfrac{5-y}{4}}$
Interchange $x$ and $y$:
$y=\sqrt[3]{\dfrac{5-x}{4}}$
The inverse of the initial function is $f^{-1}(x)=\sqrt[3]{\dfrac{5-x}{4}}$
