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

$x^{16}y^{8}$
Simplify. $(\sqrt [5]{{x^4y^2}})^{20}$ Since, $\sqrt[n] {p^m}=p^{\frac{m}{n}}$ Thus, $(\sqrt [5]{{x^4y^2}})^{20}=(x^4y^2)^{\frac{20}{5}}=(x^4y^2)^{4}$ or, $=(x^4)^{4}(y^2)^4$ or, $=x^{16}y^{8}$