Algebra: A Combined Approach (4th Edition)

$\sqrt[3]{\dfrac{2x}{27y^{12}}}=\dfrac{\sqrt[3]{2x}}{3y^{4}}$
$\sqrt[3]{\dfrac{2x}{27y^{12}}}$ Take the cubic root of both the numerator and the denominator and simplify: $\sqrt[3]{\dfrac{2x}{27y^{12}}}=\dfrac{\sqrt[3]{2x}}{\sqrt[3]{27y^{12}}}=\dfrac{\sqrt[3]{2x}}{3y^{4}}$