#### Answer

$x=\left\{ \dfrac{1}{4},1 \right\}$$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $
(2x-1)^{2/3}=x^{1/3}
,$ raise both sides to the third power to get rid of the denominator in the fractional exponent. Then express the resulting equation in the form $ax^2+bx+c=0,$ and use the concepts of factoring quadratic equations to solve the resulting equation. Finally, do checking if the solution satisfies the original equation.
$\bf{\text{Solution Details:}}$
Raising both sides to the third power, the given equation becomes \begin{array}{l}\require{cancel} \left((2x-1)^{2/3}\right)^3=\left( x^{1/3} \right)^3 .\end{array}
Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} (2x-1)^{\frac{2}{3}\cdot3 }=x^{\frac{1}{3}\cdot3} \\\\ (2x-1)^2=x .\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (2x)^2-2(2x)(1)+(1)^2=x \\\\ 4x^2-4x+1=x \\\\ 4x^2+(-4x-x)+1=0 \\\\ 4x^2-5x+1=0 .\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 4(1)=4 $ and the value of $b$ is $ -5 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -1,-4 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4x^2-x-4x+1=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} (4x^2-x)-(4x-1)=0 .\end{array}
Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(4x-1)-(4x-1)=0 .\end{array} Factoring the $GCF= (4x-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (4x-1)(x-1)=0 .\end{array}
Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 4x-1=0 \\\\\text{OR}\\\\ x-1=0 .\end{array}
Solving each equation results to \begin{array}{l}\require{cancel} 4x-1=0 \\\\ 4x=1 \\\\ x=\dfrac{1}{4} \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array}
Upon checking, $
x=\left\{ \dfrac{1}{4},1 \right\}$ satisfy the original equation.