Answer
See the explanation below.
Work Step by Step
(a)
Consider $f\left( x \right)=8{{x}^{3}}+1$
Step 1: Replace $f\left( x \right)$ with $y$:
$y=8{{x}^{3}}+1$
Step 2: Interchange $x$ and $y$:
$x=8{{y}^{3}}+1$
Step 3: Now solve for the value of $y$:
$x-1=8{{y}^{3}}$
Or
${{y}^{3}}=\frac{x-1}{8}$
That is,
$y=\sqrt[3]{\frac{x-1}{8}}$
Step 4: Replace $y$ with ${{f}^{-1}}\left( x \right)$:
${{f}^{-1}}\left( x \right)=\sqrt[3]{\frac{x-1}{8}}$
Hence, the inverse function ${{f}^{-1}}\left( x \right)$ of the $f\left( x \right)=8{{x}^{3}}+1$ is ${{f}^{-1}}\left( x \right)=\sqrt[3]{\frac{x-1}{8}}$.
(b)
Consider the function, $f\left( {{f}^{-1}}\left( x \right) \right)$
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \sqrt[3]{\frac{x-1}{8}} \right) \\
& =8{{\left( \sqrt[3]{\frac{x-1}{8}} \right)}^{3}}+1 \\
& =8\left( \frac{x-1}{8} \right)+1 \\
& =x
\end{align}$
Next consider the function, ${{f}^{-1}}\left( f\left( x \right) \right)$
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( 8{{x}^{3}}+1 \right) \\
& =\sqrt[3]{\frac{\left( 8{{x}^{3}}+1 \right)-1}{8}} \\
& =\sqrt[3]{{{x}^{3}}} \\
& =x
\end{align}$
Hence, $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and ${{f}^{-1}}\left( f\left( x \right) \right)=x$ for the function $f\left( x \right)=8{{x}^{3}}+1$.
