Answer
$\{-27,64\}$
Work Step by Step
We have to solve the quadratic equation:
$$x^{2/3}-x^{1/3}-12=0.\tag1$$
We substitute $y=x^{1/3}$:
$$y^2-y-12=0.\tag2$$
We solve equation $(2)$:
$$\begin{align*}
(y+3)(y-4)&=0\\
y+3=0&\text{ or }y-4=0\\
y=-3&\text{ or }y=4.
\end{align*}$$
We determine $x$ using the values of $y$ that we found.
$$\begin{align*}
x&=y^3\\
y&=-3\Rightarrow x=(-3)^3=-27\\
y&=4\Rightarrow x=4^3=64\\
\end{align*}$$
The solution set of the equation is:
$$\left\{-27,64\right\}.$$
