Answer
$f^{-1} (x) =\sqrt [3] {x+1}$
Work Step by Step
We have $f^{-1} (x) =\sqrt [3] {x+1}$
and $f(f^{-1} (x))=f(\sqrt [3] {x+1})=(\sqrt [3] {x+1})^3-1=x+1-1=x$
Now, $f(f^{-1} (x))=f^{-1} (x^3+1)= \sqrt [3] {(x^3+1)-1}=\sqrt[3] x^3=x$
Thus, we have $f^{-1} (x) =\sqrt [3] {x+1}$
