Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review: 14

Answer

$\frac{3y}{z^{4}}$

Work Step by Step

$\sqrt[3] (\frac{27y^{3}}{z^{12}})=\sqrt[3] (27y^{3})\times\sqrt[3] (\frac{1}{z^{12}})=3y\times\frac{1}{z^{4}}=\frac{3y}{z^{4}}$ We know that $\sqrt[3] (27y^{3})=3y$, because $(3y)^{3}=(3\times3\times3)\times y^{3}=27y^{3}$. We know that $\sqrt (\frac{1}{z^{12}})=\frac{1}{z^{4}}$, because $(\frac{1}{z^{4}})^{3}=(\frac{1}{z^{4\times3}})=\frac{1}{z^{12}}$
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