Answer
a) ${{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}$
b) they are inverses
Work Step by Step
(a)
Consider the function,
$f\left( x \right)={{x}^{3}}-1$
Step-1-Replace $f\left( x \right)$ with $y$ in $f\left( x \right)$
$\begin{align}
& f\left( x \right)={{x}^{3}}-1 \\
& y={{x}^{3}}-1
\end{align}$
Step-2-Interchange $x$ and $y$
$\begin{align}
& y={{x}^{3}}-1 \\
& x={{y}^{3}}-1 \\
\end{align}$
Step-3-Solve for $y$. Add $1$ to each side
$\begin{align}
& x={{y}^{3}}-1 \\
& x+1={{y}^{3}}-1+1 \\
& x+1={{y}^{3}}
\end{align}$
Take the cube root of each side
$\begin{align}
& x+1={{y}^{3}} \\
& \sqrt[3]{x+1}=\sqrt[3]{{{y}^{3}}} \\
& \sqrt[3]{x+1}=y
\end{align}$
Step-4-Replace $y$ in step 3 by ${{f}^{-1}}\left( x \right)$
$\begin{align}
& y=\sqrt[3]{x+1} \\
& {{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}
\end{align}$
Therefore, the required inverse of the function is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}$
(b)
Consider the functions:
$f\left( x \right)={{x}^{3}}-1$ and ${{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}$
Replace $x$ with ${{f}^{-1}}\left( x \right)$
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)={{\left( {{f}^{-1}}\left( x \right) \right)}^{3}}-1 \\
& ={{\left( \sqrt[3]{x+1} \right)}^{3}}-1
\end{align}$
Simplify
$\begin{align}
& f\left( g\left( x \right) \right)={{\left( \sqrt[3]{x+1} \right)}^{3}}-1 \\
& =\left( x+1 \right)-1 \\
& =x
\end{align}$
Now, find ${{f}^{-1}}\left( f\left( x \right) \right)$
The equation for ${{f}^{-1}}\left( x \right)$ is given as:
${{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}$
Replace $x$ with $f\left( x \right)$
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)=\sqrt[3]{f\left( x \right)+1} \\
& =\sqrt[3]{\left( {{x}^{3}}-1 \right)+1}
\end{align}$
Simplify
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)=\sqrt[3]{\left( {{x}^{3}}-1 \right)+1} \\
& =\sqrt[3]{{{x}^{3}}} \\
& =x
\end{align}$
Thus, $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and ${{f}^{-1}}\left( f\left( x \right) \right)=x$
It can be easily observed that ${{f}^{-1}}$ can be expressed in $f$ if we add 1 and then take the cube root over the whole expression.
Hence, the functions $f$ and \[{{f}^{-1}}\] are inverses of each other.
