## Precalculus (6th Edition)

$\color{blue}{\dfrac{2x\sqrt[4]{2xy^3}}{y^2}}$
Rationalize the denominator by multiplying $y^3$ to both the numerator and denominator of the radicand to obtain: $=\sqrt[4]{\dfrac{32x^5(y^3)}{y^5(y^3)}} \\=\sqrt[4]{\dfrac{32x^5y^3}{y^8}} \\=\sqrt[4]{\dfrac{32x^5y^3}{(y^2)^4}}$ Bring out the fourth root of the denominator to obtain: $\\=\dfrac{\sqrt[4]{32x^5y^3}}{y^2} \\=\dfrac{\sqrt[4]{2^5x^5y^3}}{y^2}$ Factor the radicand such that at least one factor is a perfect fourth power to obtain: $\\=\dfrac{\sqrt[4]{2^4x^4(2xy^3)}}{y^2}$ Bring out the fourth root of the perfect fourth power factor/s of the numerator to obtain: $\\=\color{blue}{\dfrac{2x\sqrt[4]{2xy^3}}{y^2}}$