## Precalculus: Mathematics for Calculus, 7th Edition

$a)$ $\sqrt[4]{x^{4}y^{2}z^{2}}=x\sqrt[4]{y^{2}z^{2}}$ $b)$ $\sqrt[3]{\sqrt{64x^{6}}}=2x$
$a)$ $\sqrt[4]{x^{4}y^{2}z^{2}}$ Take the 4th root of each factor. $\sqrt[4]{x^{4}y^{2}z^{2}}=(\sqrt[4]{x^{4}})(\sqrt[4]{y^{2}z^{2}})=x\sqrt[4]{y^{2}z^{2}}$ $b)$ $\sqrt[3]{\sqrt{64x^{6}}}$ Evaluate $\sqrt{64x^{6}}$ $\sqrt[3]{\sqrt{64x^{6}}}=\sqrt[3]{8x^{3}}=...$ Take the cubic root of each factor: $...=(\sqrt[3]{8})(\sqrt[3]{x^{3}})=2x$