Answer
$\dfrac{14x\sqrt[3]{2x}}{9}$
Work Step by Step
Using the properties of radicals, then,
\begin{array}{l}
-\dfrac{\sqrt[3]{2x^4}}{9}+\sqrt[3]{\dfrac{250x^4}{27}}
\\\\=
-\dfrac{\sqrt[3]{x^3\cdot2x}}{9}+\sqrt[3]{\dfrac{125x^3}{27}\cdot 2x}
\\\\=
-\dfrac{x\sqrt[3]{2x}}{9}+\dfrac{5x\sqrt[3]{2x}}{3}
\\\\=
\dfrac{-x\sqrt[3]{2x}+3(5x\sqrt[3]{2x})}{9}
\\\\=
\dfrac{-x\sqrt[3]{2x}+15x\sqrt[3]{2x}}{9}
\\\\=
\dfrac{14x\sqrt[3]{2x}}{9}
.\end{array}
