Answer
$\left\{-8,1\right\}$
Work Step by Step
Let $z=
x^{1/3}
$. Then the given equation, $
x^{2/3}+x^{1/3}-2=0
,$ is equivalent to
\begin{align*}
\left(x^{1/3}\right)^2+x^{1/3}-2=0
\\
z^2+z-2=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z-1)(z+2)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z-1=0 & z+2=0
\\
z=1 & z=-2
.\end{array}
Since $z=
x^{1/3}
,$ by back substitution, then
\begin{array}{l|r}
x^{1/3}=1 & x^{1/3}=-2
\\
\left(x^{1/3}\right)^3=(1)^3 & \left(x^{1/3}\right)^3=(-2)^3
\\
x=1 & x=-8
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=1: & \text{If }x=-8
\\\\
1^{2/3}+1^{1/3}-2\overset{?}=0 &
(-8)^{2/3}+(-8)^{1/3}-2\overset{?}=0
\\
1+1-2\overset{?}=0 &
\left(\sqrt[3]{-8}\right)^2+\left(\sqrt[3]{-8}\right)^1-2\overset{?}=0
\\
0\overset{\checkmark}=0 &
\left(-2\right)^2+\left(-2\right)^1-2\overset{?}=0
\\
&
4-2-2\overset{?}=0
\\
&
0\overset{\checkmark}=0
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
x^{2/3}+x^{1/3}-2=0
$ is $\left\{-8,1\right\}$.
