#### Answer

$x=\left\{ \dfrac{4}{3},2 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt[3]{3x^2-9x+8}=\sqrt[3]{x}
,$ cube both sides of the equal sign. Then use the properties of equality to express the resulting equation in the form $ax^2+bx+c=0.$ Next is to factor the quadratic equation and solve for the value/s of the variable using the Zero Product Property.
$\bf{\text{Solution Details:}}$
Raising both sides to the third power and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
3x^2-9x+8=x
\\\\
3x^2+(-9x-x)+8=0
\\\\
3x^2-10x+8=0
.\end{array}
In the trinomial expression above, the value of $ac$ is $
3(8)
$ and the value of $b$ is $
-10
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-4,-6
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
3x^2-4x-6x+8=0
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(3x^2-4x)-(6x-8)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(3x-4)-2(3x-4)=0
.\end{array}
Factoring the $GCF=
(3x-4)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3x-4)(x-2)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
3x-4=0
\\\\\text{OR}\\\\
x-2=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
3x-4=0
\\\\
3x=4
\\\\
x=\dfrac{4}{3}
\\\\\text{OR}\\\\
x-2=0
\\\\
x=2
.\end{array}
Hence, $
x=\left\{ \dfrac{4}{3},2 \right\}
.$