Answer
The solutions are $x=-\frac{27}{8}$ and $x=-\frac{125}{8}$.
Work Step by Step
Add $15$ to both sides of the equation:
$4x^{2/3}+16x^{1/3}+15=0
\\4(x^{1/3})^2+16x^{1/3}+15=0$
Let $u=x^{1/3}$.
Rewrite the equation above using $u$ to obtain:
$4u^2+16u+15=0$
Factor the trinomial to obtain:
$(2u+3)(2u+5)=0$
Factor each factor with a variable to zero then solve each equation to obtain:
\begin{array}{ccc}
&2u+3=0 &\text{or} &2u+5=0
\\&2u=-3 &\text{or} &2u=-5
\\&u=-\frac{3}{2} &\text{or} &u=-\frac{5}{2}
\end{array}
Since $u=x^{1/3}$, then:
\begin{array}{ccc}
&u=-\frac{3}{2} &\text{or} &u=-\frac{5}{2}
\\&x^{1/3}=-\frac{3}{2} &\text{or} &x^{1//3}=-\frac{5}{2}
\\&(x^{1/3})^3=(-\frac{3}{2})^3 &\text{or} &(x^{1//3})^3=(-\frac{5}{2})^3
\\&x=-\frac{27}{8} &\text{or} &x=-\frac{125}{8}
\end{array}
Therefore, the solutions are $x=-\frac{27}{8}$ and $x=-\frac{125}{8}$.
