## Intermediate Algebra for College Students (7th Edition)

$\dfrac{3x^2\sqrt[3]{3x^2}}{2y^5}$
RECALL: The quotient rule: $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ where $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers and $b\ne0$ Use the quotient rule above to obtain: $=\dfrac{\sqrt[3]{81x^8}}{\sqrt[3]{8y^{15}}}$ Factor each radicand so that at least one factor is a perfect cube to obtain: $=\dfrac{\sqrt[3]{27x^6(3x^2)}}{\sqrt[3]{(2y^5)^3}} \\=\dfrac{\sqrt[3]{(3x^2)^3(3x^2)}}{\sqrt[3]{(2y^5)^3}}$ Simplify each radical to obtain; $=\dfrac{3x^2\sqrt[3]{3x^2}}{2y^5}$