Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.2 - Rational Exponents - Exercise Set - Page 522: 89

Answer

$\sqrt[3]{4a^2}$

Work Step by Step

RECALL: (i) $\sqrt[n]{a} = a^{\frac{1}{n}}$ (ii) $\sqrt[n]{a^m} = a^{\frac{m}{n}}$ (iii) $(ab)^m = a^mb^m$ (iv) $(a^m)^n=a^{mn}$ Use the rules above to obtain: $=\left((2a)^{\frac{1}{6}}\right)^4 \\=(2a)^{\frac{1}{6} \cdot 4} \\=(2a)^{\frac{2}{3}} \\=\sqrt[3]{(2a)^2} \\=\sqrt[3]{4a^2}$
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