Answer
$\left\{-1,27\right\}$
Work Step by Step
Let $z=
x^{1/3}
$. Then the given equation, $
x^{2/3}-2x^{1/3}-3=0
,$ is equivalent to
\begin{align*}
\left(x^{1/3}\right)^2-2x^{1/3}-3=0
\\
z^2-2z-3=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-3)(z+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-3=0 & z+1=0
\\
z=3 & z=-1
.\end{array}
Since $z=
x^{1/3}
,$ by back substitution, then
\begin{array}{l|r}
x^{1/3}=3 & x^{1/3}=-1
\\
\left(x^{1/3}\right)^3=(3)^3 & \left(x^{1/3}\right)^3=(-1)^3
\\
x=27 & x=-1
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=27: & \text{If }x=-1
\\\\
27^{2/3}-2(27)^{1/3}-3\overset{?}=0 &
(-1)^{2/3}-2(-1)^{1/3}-3\overset{?}=0
\\
\left(\sqrt[3]{27}\right)^2-2\left(\sqrt[3]{27}\right)^1-3\overset{?}=0 &
\left(\sqrt[3]{-1}\right)^2-2\left(\sqrt[3]{-1}\right)^1-3\overset{?}=0
\\
\left(3\right)^2-2\left(3\right)^1-3\overset{?}=0 &
\left(-1\right)^2-2\left(-1\right)^1-3\overset{?}=0
\\
9-6-3\overset{?}=0 &
1+2-3\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 $
x^{2/3}-2x^{1/3}-3=0
$ is $\left\{-1,27\right\}$.
