Answer
$x=0$ and $x=2$
Work Step by Step
$\sqrt[3]{4x^{2}-4x}=x$
Cube both sides:
$(\sqrt[3]{4x^{2}-4x})^{3}=x^{3}$
$4x^{2}-4x=x^{3}$
Take all terms to the right:
$0=x^{3}-4x^{2}+4x$
Reorganize:
$x^{3}-4x^{2}+4x=0$
Take out common factor $x$:
$x(x^{2}-4x+4)=0$
The trinomial inside the parentheses is a perfect square trinomial. Factor it:
$x(x-2)^{2}=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x=0$
$(x-2)^{2}=0$
$\sqrt{(x-2)^{2}}=\sqrt{0}$
$x-2=0$
$x=2$
The solutions found are $x=0$ and $x=2$. The original equation is true for both of these values. The final answer is:
$x=0$ and $x=2$
