#### Answer

The solutions are $x=\frac{1}{27}$ and $x=-64$.

#### Work Step by Step

Subtract $4$ to both sides of the equation to obtain:
$3x^{2/3}+11x^{1/3}-4=0
\\3(x^{1/3})^2+11x^{1/3}-4=0$
Let
$u=x^{1/3}$
Rewrite the equation above using the variable $u$ to obtain:
$3u^2+11u-4=0$
Factor the trinomial to obtain:
$(3u-1)(u+4)=0$
Use the Zero-Factor Property by equating each unique factor to zero. then solve each equation:
\begin{array}{ccc}
&3u-1 = 0 &\text{ or } &u+4=0
\\&3u=1 &\text{ or } &u=-4
\\&u=\frac{1}{3} &\text{ or } &u=-4
\end{array}
Since $u=x^{1/3}$, then
\begin{array}{ccc}
&u= \frac{1}{3} &\text{or} &u=-4
\\&x^{1/3}=\frac{1}{3} &\text{or} &x^{1/3}=-4
\\&(x^{1/3})^3=(\frac{1}{3})^3 &\text{or} &(x^{1/3})^3=(-4)^3
\\&x=\frac{1}{27} &\text{or} &x=-64
\end{array}
Thus, the solutions are $x=\frac{1}{27}$ and $x=-64$.