Answer
$\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{9x^{2}}}=\dfrac{\sqrt[3]{6xy^{2}}}{3x}$
Work Step by Step
$\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{9x^{2}}}$
Multiply the fraction by $\dfrac{\sqrt[3]{3x}}{\sqrt[3]{3x}}$ and simplify:
$\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{9x^{2}}}=\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{9x^{2}}}\cdot\dfrac{\sqrt[3]{3x}}{\sqrt[3]{3x}}=\dfrac{\sqrt[3]{6xy^{2}}}{\sqrt[3]{27x^{3}}}=\dfrac{\sqrt[3]{6xy^{2}}}{3x}$
