College Algebra, Chapter 1, 1.5, Section 1.5, Problem 48

Find all real solutions of the equation $\displaystyle 2(x - 4)^{\frac{7}{3}} - (x - 4)^{\frac{4}{3}} - (x - 4)^{\frac{1}{3}} = 0$


$
\begin{equation}
\begin{aligned}

2(x - 4)^{\frac{7}{3}} - (x - 4)^{\frac{4}{3}} - (x - 4)^{\frac{1}{3}} =& 0
&& \text{Given}
\\
\\
2w^7 - w^4 - w =& 0
&& \text{Let } w = (x - 4)^{\frac{1}{3}}
\\
\\
w(2w^6 - w^3 - 1) =& 0
&& \text{Factor out } w
\\
\\
w =& 0 \text{ and } 2w^6 - w^3 - 1 = 0
&& \text{Zero Product Property}
\\
\\
(x - 4)^{\frac{1}{3}} =& 0
&& \text{Substitute } w = (x - 4)^{\frac{1}{3}} \text{ in } w = 0
\\
\\
x =& 4
&& \text{Solve for } x

\end{aligned}
\end{equation}
$


To find the solution in $2w^6 - w^3 - 1 = 0$,


$
\begin{equation}
\begin{aligned}

\text{We let } z =& w^3, \text{ so}
&&
\\
\\
2w^6 - w^3 - 1 =& 0
&&
\\
\\
3z^2 - z - 1 =& 0
&& \text{Factor}
\\
\\
(z - 1)(2z + 1) =& 0
&& \text{Zero Product Property}
\\
\\
z - 1 =& 0 \text{ and } 2z + 1 = 0
&& \text{Solve for } z
\\
\\
z =& 1 \text{ and } z = \frac{-1}{2}
&& \text{Substitute } z = w^3
\\
\\
w^3 =& 1 \text{ and } w^3 = \frac{-1}{2}
&& \text{Solve for } w
\\
\\
w =& 1 \text{ and } w = \frac{-1}{\sqrt[3]{2}}
&& \text{Substitute } w = (x - 4)^{\frac{1}{3}}
\\
\\
(x - 4)^{\frac{1}{3}} =& 1 \text{ and } (x - 4)^{\frac{1}{3}} = \frac{1}{3 \sqrt{2}}
&& \text{Solve for } x
\\
\\
x =& 5 \text{ and } x = \frac{-1}{2} + 4 = \frac{7}{2}
&& \text{Simplify}
\\
\\
x =& 4, x = 5 \text{ and } x = \frac{7}{2}
&& \text{are the solutions for the equation}

\end{aligned}
\end{equation}
$