#### Answer

The solutions are $x=-\frac{1}{125}$ and $x=\frac{1}{8}$.

#### Work Step by Step

The given equation can be written as:
$20(x^{1/3})^2-6x^{1/3}-2=0$
Let $u=x^{1/3}$.
Rewrite the equation above using $u$ to obtain:
$20u^2-6u-2=0$
Factor out $2$ to obtain:
$2(10u^2-3u-1)=0$
Factor the trinomial to obtain:
$2(5u+1)(2u-1)=0$
Factor each factor with a variable to zero then solve each equation to obtain:
\begin{array}{ccc}
&5u+1=0 &\text{or} &2u-1=0
\\&5u=-1 &\text{or} &2u=1
\\&u=-\frac{1}{5} &\text{or} &u=\frac{1}{2}
\end{array}
Since $u=x^{1/3}$, then:
\begin{array}{ccc}
&u=-\frac{1}{5} &\text{or} &u=\frac{1}{2}
\\&x^{1/3}=-\frac{1}{5} &\text{or} &x^{1//3}=\frac{1}{2}
\\&(x^{1/3})^3=(-\frac{1}{5})^3 &\text{or} &(x^{1//3})^3=(\frac{1}{2})^3
\\&x=-\frac{1}{125} &\text{or} &x=\frac{1}{8}
\end{array}
Therefore, the solutions are $x=-\frac{1}{125}$ and $x=\frac{1}{8}$.