Answer
$\dfrac{\sqrt[3]{3x}}{\sqrt[3]{4y^{4}}}=\dfrac{\sqrt[3]{6xy^{2}}}{2y^{2}}$
Work Step by Step
$\dfrac{\sqrt[3]{3x}}{\sqrt[3]{4y^{4}}}$
Simplify the denominator:
$\dfrac{\sqrt[3]{3x}}{\sqrt[3]{4y^{4}}}=\dfrac{\sqrt[3]{3x}}{y\sqrt[3]{4y}}=...$
Multiply the fraction by $\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{2y^{2}}}$ and simplify the expression:
$...=\dfrac{\sqrt[3]{3x}}{y\sqrt[3]{4y}}\cdot\dfrac{\sqrt[3]{2y^{2}}}{\sqrt[3]{2y^{2}}}=\dfrac{\sqrt[3]{6xy^{2}}}{y\sqrt[3]{8y^{3}}}=\dfrac{\sqrt[3]{6xy^{2}}}{y(2y)}=\dfrac{\sqrt[3]{6xy^{2}}}{2y^{2}}$
