Answer
$f^{-1}(x)=\sqrt[3]{x+1}$
(the inverse is blue on the graph)
Work Step by Step
Step 1: Replace $f(x)$ with $y$.
$y=x^{3}-1$
Step 2: Interchange $x$ and $y$.
$x=y^{3}-1$
Step 3: Solve the equation for $y$.
$ x=y^{3}-1,\qquad$ ... add 1
$ x+1=y^{3},\qquad$ ... raise to the power $(...)^{1/3}$,
$(x+1)^{1/3}=y$
$y=\sqrt[3]{x+1}$
Step 4: Replace y with the notation $f^{-1}(x)$.
$f^{-1}(x)=\sqrt[3]{x+1}$
Graphing $f(x)=x^{3}, \left[\begin{array}{lll}
x & f(x) & (x,y)\\
0 & -1 & (0,-1)\\
-1 & -2 & (-1,-2)\\
1 & 0 & (1,0)\\
2 & 7 & (2,7)
\end{array}\right], $
the graph of $f(x)$ is a smooth curve passing through the points $(x,y)$.
The graph of $f^{-1}(x)$ is a smooth curve passing through points (y,x) of the above table,
