Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.2 - Rational Exponents - 7.2 Exercises: 13

Answer

9

Work Step by Step

As shown on page 445, we know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$. Therefore, $729^{\frac{1}{3}}=\sqrt[3] 729^{1}=\sqrt[3] 729=9$. We know that $\sqrt[3] 729=9$, because $9^{3}=729$.
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