Answer
$x=0$ or $x=3$ or $x=\frac{5}{2}$
Work Step by Step
We solve by factoring:
$x(x^{2}-3x)^{1/3}+2(x^{2}-3x)^{4/3}=0$
$(x^{2}-3x)^{1/3}[x+2(x^{2}-3x)]=0$
$(x^{2}-3x)^{1/3}(x+2x^{2}-6x)=0$
$(x^{2}-3x)^{1/3}(2x^{2}-5x)=0$
$(x^{2}-3x)^{1/3}=0$ or $2x^{2}-5x=0$
$x^{2}-3x=0^3$ or $2x^{2}-5x=0$
$x(x-3)=0$ or $x(2x-5)=0$
$x=0$ or $x=3$ or $x=0$ or $2x-5=0$
$x=0$ or $x=3$ or $2x=5$
$x=0$ or $x=3$ or $x=\frac{5}{2}$