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: 17



Work Step by Step

$\sqrt[3] (8x^{6})=(8x^{6})^{\frac{1}{3}}=8^{\frac{1}{3}}\times(x^{6})^{\frac{1}{3}}=8^{\frac{1}{3}}\times(x^{\frac{6}{3}})=8^{\frac{1}{3}}\times x^{2}=2x^{2}$ We know that $8^{\frac{1}{3}}=2$, because $2^{3}=8$.
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