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.