College Algebra, Chapter 1, 1.1, Section 1.1, Problem 10

Check whether a.) $x = 4$ or b.) $x = 8$ is a solution of the equation $\displaystyle \frac{x^{\frac{3}{2}}}{x-6} = x -8$

a.) $x = 4$

$
\begin{equation}
\begin{aligned}
\frac{(4)^{\frac{3}{2}}}{4-6} &= 4 - 8 && \text{Substitute } x = 4\\
\\
\frac{\left[ (4)^{\frac{1}{2}}\right]^3}{-2} &= -4 && \text{Simplify} \\
\\
- \frac{(2)^3}{2} &= -4 && \text{Simplify} \\
\\
- \frac{8}{2} &= -4 && \text{Simplify} \\
\\
-4 &= -4
\end{aligned}
\end{equation}
$

So $x = 4$ is the solution to the equation.

b.) $x = 8$

$
\begin{equation}
\begin{aligned}
\frac{(8)^{\frac{3}{2}}}{8-6} &= 8-8 && \text{Substitute } x =8\\
\\
\frac{(4 \cdot 2)^{\frac{3}{2}}}{2} &= 0 && \text{Get the factor}\\
\\
\frac{(4)^{\frac{3}{2}} (2)^{\frac{3}{2}} }{2} &= 0 && \text{Group}\\
\\
\frac{\left[(4)^{\frac{1}{2}}\right]^3\left[(2)^3\right]^{\frac{1}{2}}}{2} &= 0 && \text{Simplify}\\
\\
\frac{(2)^3 (8)^{\frac{1}{2}}}{2} &= 0 && \text{Simplify}\\
\\
\frac{(8) (4\cdot 2)^{\frac{1}{2}}}{2} &= 0 && \text{Simplify}\\
\\
\frac{(8)(\cancel{2})(2)^{\frac{1}{2}}}{\cancel{2}} &= 0 && \text{Cancel out like terms}\\
\\
8 (2)^{\frac{1}{2}} &\neq 0
\end{aligned}
\end{equation}
$

So $x = 8$ is not the solution to the equation.