Answer
$f^{-1}(x)=\frac{1}{3}x^3+2$
Work Step by Step
1. Given $f(x)=\sqrt[3] {3x-6}$, rewrite it as $y=\sqrt[3] {3x-6}$
2. Exchange $x,y$ to get $x=\sqrt[3] {3y-6}$
3. Solve for $y$, we have $3y-6=x^3$, thus $y=\frac{1}{3}x^3+2$
4. Replace $y$ with $f^{-1}(x)$, we have $f^{-1}(x)=\frac{1}{3}x^3+2$
