Answer
The solutions are $-\frac{1}{8}$ and $-8$.
Work Step by Step
The given equation can be written as:
$2(x^{1/3})^2+3x^{1/3}-2=0$
Let $u=x^{1/3}$.
Rewrite the equation above using $u$ to obtain:
$2u^2+3u-2=0$
Factor the trinomial to obtain:
$(2u-1)(u+2)=0$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
\\&2u-1=0 &\text{or} &u+2=0
\\&2u=1 &\text{or} &u=-2
\\&u=\frac{1}{2} &\text{or} &u=-2
\end{array}
Since $u=x^{1/3}$, then:
\begin{array}{ccc}
\\&u=\frac{1}{2} &\text{or} &u=-2
\\&x^{1/3}=\frac{1}{2} &\text{or} &x^{1/3}=-2
\\&(x^{1/3})^3=(\frac{1}{2})^3 &\text{or} &(x^{1/3})^3=(-2)^3
\\&x=-\frac{1}{8} &\text{or} &x=-8
\end{array}
Therefore, the solutions are $-\frac{1}{8}$ and $-8$.
