Answer
$\left\{-\dfrac{512}{27},27\right\}$
Work Step by Step
Let $z=
x^{1/3}
$. Then the given equation, $
3x^{2/3}-x^{1/3}-24=0
,$ is equivalent to
\begin{align*}
3\left(x^{1/3}\right)^2-x^{1/3}-24&=0
\\
3z^2-z-24&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-3)(3z+8)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-3=0 & 3z+8=0
\\
z=3 & 3z=-8
\\\\
& z=-\dfrac{8}{3}
.\end{array}
Since $z=
x^{1/3}
,$ by back substitution, then
\begin{array}{l|r}
x^{1/3}=3 & x^{1/3}=-\dfrac{8}{3}
\\\\
\left(x^{1/3}\right)^3=(3)^3 & \left(x^{1/3}\right)^3=\left(-\dfrac{8}{3}\right)^3
\\\\
x=27 & x=-\dfrac{512}{27}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=27: & \text{If }x=-\dfrac{512}{27}
\\\\
3(27)^{2/3}-(27)^{1/3}-24\overset{?}=0 &
3\left(-\dfrac{512}{27}\right)^{2/3}-\left(-\dfrac{512}{27}\right)^{1/3}-24\overset{?}=0
\\\\
3\left(\sqrt[3]{27}\right)^2-\left(\sqrt[3]{27}\right)^1-24\overset{?}=0 &
3\left(\sqrt[3]{-\dfrac{512}{27}}\right)^2-\left(\sqrt[3]{-\dfrac{512}{27}}\right)^1-24\overset{?}=0
\\\\
3\left(3\right)^2-\left(3\right)^1-24\overset{?}=0 &
3\left(-\dfrac{8}{3}\right)^2-\left(-\dfrac{8}{3}\right)^1-24\overset{?}=0
\\\\
3\left(9\right)-3-24\overset{?}=0 &
3\left(\dfrac{64}{9}\right)+\dfrac{8}{3}-24\overset{?}=0
\\\\
27-3-24\overset{?}=0 &
\dfrac{64}{3}+\dfrac{8}{3}-24\overset{?}=0
\\\\
0\overset{\checkmark}=0 &
\dfrac{72}{3}-24\overset{?}=0
\\\\
&
24-24\overset{?}=0
\\
&
0\overset{\checkmark}=0
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
3x^{2/3}-x^{1/3}-24=0
$ is $\left\{-\dfrac{512}{27},27\right\}$.
