#### Answer

The solutions are $x=-\frac{1}{8}$ and $x=27$.

#### Work Step by Step

Subtract $3$ to both sides of the equation to obtain:
$2x^{2/3}-5x^{1/3}-3=0
\\2(x^{1/3})^2-5x^{1/3}-3=0$
Let
$u=x^{1/3}$
Rewrite the equation above using the variable $u$ to obtain:
$2u^2-5u-3=0$
Factor the trinomial to obtain:
$(2u+1)(u-3)=0$
Use the Zero-Factor Property by equating each unique factor to zero. then solve each equation:
\begin{array}{ccc}
&2u+1 = 0 &\text{ or } &u-3=0
\\&2u=-1 &\text{ or } &u=3
\\&u=-\frac{1}{2} &\text{ or } &u=3
\end{array}
Since $u=x^{1/3}$, then
\begin{array}{ccc}
&u= -\frac{1}{2} &\text{or} &u=3
\\&x^{1/3}=-\frac{1}{2} &\text{or} &x^{1/3}=3
\\&(x^{1/3})^3=(-\frac{1}{2})^3 &\text{or} &(x^{1/3})^3=3^3
\\&x=-\frac{1}{8} &\text{or} &x=27
\end{array}
Thus, the solutions are $x=-\frac{1}{8}$ and $x=27$.