Answer
The solutions are $x=-\frac{3}{2}$ and $x=-\frac{3}{4}$
Work Step by Step
$\bigg(\dfrac{1}{x+1}\bigg)^2-2\bigg(\dfrac{1}{x+1}\bigg)-8=0\hspace{0.4cm} \color{blue}{\text{Given equation}}$
$W^2-2W-8=0\hspace{3.3cm} \color{blue}{\text{Let } W=\dfrac{1}{x+1}}$
$(W+2)+(W-4)=0\hspace{2.3cm} \color{blue}{\text{Factor}}$
$W+2=0\hspace{0.5cm}or\hspace{0.5cm}W-4=0\hspace{0.9cm} \color{blue}{\text{Zero-Product Property}}$
$W=-2\hspace{1.5cm}W=4\hspace{1.99cm} \color{blue}{\text{Solve}}$
Now change $W$ back into the correspondimg values of $x$
$\dfrac{1}{x+1}=-2\hspace{1.5cm}\dfrac{1}{x+1}=4$
$x+1=-\frac{1}{2}\hspace{1.5cm}x+1=\frac{1}{4}$
$x=-\frac{1}{2}-1\hspace{1.5cm}x=\frac{1}{4}-1$
$x=-\frac{3}{2}\hspace{1.5cm}x=-\frac{3}{4}$
The solutions are $x=-\frac{3}{2}$ and $x=-\frac{3}{4}$
