Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Review Exercises - Page 143: 59


$ y\cdot \sqrt[3]{y^{2}}$

Work Step by Step

We apply $\qquad \sqrt[n]{y^{n}}=y,\qquad $ when n is odd. $\sqrt[3]{y^{5}}= \sqrt[3]{y^{3}\cdot y^{2}}$ Apply $\sqrt[n]{a\cdot b}$ = $\sqrt[n]{a}\cdot \sqrt[n]{b}$ $= \sqrt[3]{y^{3}}\cdot \sqrt[3]{y^{2}} $ $=y \cdot \sqrt[3]{y^{2}}$
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