#### Answer

$x=\left\{ \dfrac{2}{5},1 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $ \sqrt[3]{5x^2-6x+2}-\sqrt[3]{x}=0 ,$ transpose the second term and then 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.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel}
\sqrt[3]{5x^2-6x+2}=\sqrt[3]{x} .\end{array}
Raising both sides to the third power and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel}
5x^2-6x+2=x \\\\
5x^2+(-6x-x)+2=0
\\\\
5x^2-7x+2=0
.\end{array}
In the trinomial expression above, the value of $ac$ is $ 5(2)=10 $ and the value of $b$ is $ -7 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -2,-5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel}
5x^2-2x-5x+2=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}
(5x^2-2x)-(5x-2)=0
.\end{array}
Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel}
x(5x-2)-(5x-2)=0
.\end{array}
Factoring the $GCF= (5x-2) $ of the entire expression above results to \begin{array}{l}\require{cancel}
(5x-2)(x-1)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 5x-2=0 \\\\\text{OR}\\\\ x-1=0 .\end{array}
Solving each equation results to \begin{array}{l}\require{cancel} 5x-2=0 \\\\ 5x=2 \\\\ x=\dfrac{2}{5} \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array}
Hence, $
x=\left\{ \dfrac{2}{5},1 \right\}
.$