#### Answer

$x=4$ or $x=1$

#### Work Step by Step

We are given:
$(x-2)^{2/3}=x^{1/3}$
We raise both sides to the 3rd power:
$[(x-2)^{2/3}]^{3}=(x^{1/3})^{3}$
$(x-2)^{2}=x^1$
$(x-2)(x-2)=x$
And distribute:
$x^{2}-4x+4=x$
$x^{2}-5x+4=0$
And factor:
$(x-4)(x-1)=0$
Equate each factor to zero, then solve each equation for $x$:
$(x-4)=0$ or $(x-1)=0$
$x=4$ or $x=1$