## Precalculus (6th Edition)

$\dfrac{\sqrt[3]{36p^2}}{4p^2}$
Rationalize the denominator by multiplying $4p^2$ to both the numerator and denominator of the radicand to obtain: $=\sqrt[3]{\dfrac{9(4p^2)}{16p^4(4p^2)}} \\=\sqrt[3]{\dfrac{36p^2}{64p^6}} \\=\sqrt[3]{\dfrac{36p^2}{(4p^2)^3}}$ Bring out the cube root of the denominator to obtain: $\\=\color{blue}{\dfrac{\sqrt[3]{36p^2}}{4p^2}}$