Answer
$\left\{-64,27\right\}$
Work Step by Step
Let $z=
r^{1/3}
$. Then the given equation, $
r^{2/3}+r^{1/3}-12=0
,$ is equivalent to
\begin{align*}
\left(r^{1/3}\right)^2+r^{1/3}-12=0
\\
z^2+z-12=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z+4)(z-3)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z+4=0 & z-3=0
\\
z=-4 & z=3
.\end{array}
Since $z=
r^{1/3}
,$ by back substitution, then
\begin{array}{l|r}
r^{1/3}=-4 & r^{1/3}=3
\\
\left(r^{1/3}\right)^3=(-4)^3 & \left(r^{1/3}\right)^3=(3)^3
\\
r=-64 & r=27
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }r=-64: & \text{If }r=27
\\\\
(-64)^{2/3}+(-64)^{1/3}-12\overset{?}=0 &
27^{2/3}+27^{1/3}-12\overset{?}=0
\\
\left(\sqrt[3]{-64}\right)^2+\left(\sqrt[3]{-64}\right)^1-12\overset{?}=0 &
\left(\sqrt[3]{27}\right)^2+\left(\sqrt[3]{27}\right)^1-12\overset{?}=0
\\
\left(-4\right)^2+\left(-4\right)^1-12\overset{?}=0 &
\left(3\right)^2+\left(3\right)^1-12\overset{?}=0
\\
16-4-12\overset{?}=0 &
9+3-12\overset{?}=0
\\
0\overset{\checkmark}=0 &
0\overset{\checkmark}=0
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
r^{2/3}+r^{1/3}-12=0
$ is $\left\{-64,27\right\}$.
