Show Algebraically That the Functions $f$ And $g$ Shown Below Are Inverse Functions

Multiple Choice

Show algebraically that the functions $f$ and $g$ shown below are inverse functions.
$f ( x ) = \sqrt [ 3 ] { 8 x - 7 } , \ g ( x ) = \frac { x ^ { 3 } + 7 } { 8 }$

A)
$\begin{aligned}f(g(x) ) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right) -7} \\&=\sqrt[3]{\left(x^{3}+56\right) -56} \\&=\sqrt[3]{x^{3}+56-56} \\&=\sqrt[3]{x^{3}} \\&=x\end{aligned}$


$\begin{aligned}g(f(x) ) &=\frac{(\sqrt[3]{8 x-7}) ^{3}+7}{8} \\&=\frac{8 x-7^{3}+7^{3}}{8} \\&=\quad \frac{8 x}{8}\\&=x\end{aligned}$

B) $\begin{aligned}f(g(x) ) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right) -7} \\&=\sqrt[3]{\left(x^{3}+7\right) -7} \\&=\sqrt[3]{x^{3}+7-7} \\&=\sqrt[3]{x^{3}} \\&=x\end{aligned}$

$\begin{aligned}g(f(x) ) &=& \frac{(\sqrt[3]{8 x-7}) ^{3}+7}{8} \\&=& \frac{8 x-7+7}{8} \\&=& \frac{8 x}{8} \\&=& x\end{aligned}$
C) $\begin{aligned}f(g(x) ) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right) -7} \\&=\sqrt[3]{\left(\frac{8 x^{3}+7}{8}\right) -7} \\&=\sqrt[3]{x^{3}+7-7} \\&=\sqrt[3]{x^{3}} \\&=x\end{aligned}$

$\begin{aligned}g(f(x) ) &=\frac{(\sqrt[3]{8 x-7}) ^{3}+7}{8} \\&=\frac{8^{3} x-7+7}{8^{3}} \\&=\frac{8^{3} x}{8^{3}} \\&=x\end{aligned}$
D) $\begin{aligned}f(g(x) ) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right) -7} \\&=\sqrt[3]{\left(8 x^{3}+56\right) -56} \\&=\sqrt[3]{8 x^{3}+56-56} \\&=\sqrt[3]{8 x^{3}} \\&=x\end{aligned}$

$\begin{aligned}g(f(x) ) &=\frac{(\sqrt[3]{8 x-7}) ^{3}+7}{8} \\&=\frac{8^{3} x-7^{3}+7^{3}}{8^{3}} \\&=\frac{8^{3} x}{8^{3}} \\&=\quad x\end{aligned}$
E) $\begin{aligned}f(g(x) ) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right) -7} \\&=\sqrt[3]{\left(x^{3}+\frac{7}{8}\right) -7} \\&=\sqrt[3]{x^{3}+\frac{0}{8}} \\&=\sqrt[3]{x^{3}} \\&=x\end{aligned}$

$\begin{aligned}g(f(x) ) &=\frac{(\sqrt[3]{8 x-7}) ^{3}+7}{8} \\&=\frac{24 x-21+21}{24} \\&=\quad \frac{24 x}{24} \\&=\quad x\end{aligned}$