Intermediate Algebra (6th Edition)

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

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents: 26



Work Step by Step

$\sqrt[3] (8x)\times\sqrt[3] (8x^{2})=\sqrt[3] (8x\times8x^{2})=\sqrt[3] (64x^{3})=\sqrt[3] (64\times x^{3})=\sqrt[3] 64\times\sqrt[3] (x^{3})=4x$ We know that $\sqrt[3] 64=4$, because $4^{3}=64$. We know that $\sqrt[3] (x^{3})=x$, because $(x)^{3}=x^{3}$.
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