## College Algebra 7th Edition

$\dfrac{3y}{x}$
RECALL: (i) $a^{-m} = \dfrac{1}{a^m},a\ne0$ (ii) $\sqrt[n]{a^m} = a^{\frac{m}{n}}$ (iii) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$ Simplify the radicand by canceling common factors: $\require{cancel} =\sqrt[3]{\dfrac{\cancel{54}27\cancel{x^2}\cancel{y^4}y^3}{\cancel{2}\cancel{x^5}x^3\cancel{y}}} \\=\sqrt[3]{\dfrac{27y^3}{x^3}}$ Factor the radicand so that at least one factor is a perfect cube to obtain: $\\=\sqrt[3]{\dfrac{3^3y^3}{x^3}}$ Simplify to obtain: $=\dfrac{3y}{x}$