Answer
$\left\{-1, \dfrac{1}{2}\right\}$
Work Step by Step
Let $u=x^{-3}$.
Then, the given equation becomes:
$$u^2-7u-8=0$$
Factor the trinomial to obtain:
$$(u-8)(u+1)=0$$
Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain:
\begin{align*}
u-8&=0 &\text{or}& &u+1=0\\
u&=8 &\text{or}& &u=-1
\end{align*}
With $u=x^{-3}$, substitute $x^{-3}$ to $u$ to obtain:
\begin{align*}
x^{-3}&=8 &\text{or}& &x^{-3}=-1\\
\frac{1}{x^3}&=8 &\text{or}& &\frac{1}{x^3}=-1\\
\frac{1}{x^3}&=8 &\text{or}& &\frac{1}{x^3}=-1\\
8x^3&=1 &\text{or}& &-x^3=1\\
x^3&=\frac{1}{8} &\text{or}& &x^3=-1\\
\sqrt[3]{x^3}&=\sqrt[3]{\frac{1}{8}} &\text{or}& &\sqrt[3]{x^3}=\sqrt[3]{-1}\\
x&=\frac{1}{2} &\text{or}& &x=-1
\end{align*}