Answer
$f^{-1}(x)=\sqrt[3]{2-5x}$
Work Step by Step
$f(x)=\dfrac{2-x^{3}}{5}$
Rewrite this expression as $y=\dfrac{2-x^{3}}{5}$ and solve for $x$:
$y=\dfrac{2-x^{3}}{5}$
Take the $5$ to multiply the left side:
$5y=2-x^{3}$
Take $-x^{3}$ to the left side:
$5y+x^{3}=2$
Take $5y$ to the right side:
$x^{3}=2-5y$
Take the cubic root of both sides:
$\sqrt[3]{x^{3}}=\sqrt[3]{2-5y}$
$x=\sqrt[3]{2-5y}$
Interchange $x$ and $y$:
$y=\sqrt[3]{2-5x}$
The inverse of the initial function is $f^{-1}(x)=\sqrt[3]{2-5x}$