Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises: 57

Answer

$\color{blue}{32\sqrt[3]{2}}$

Work Step by Step

Note that $(-2)^4=16$ and $16=2^4$. Thus, the given expression is equivalent to: $=\sqrt[3]{2^4(16)(2^8)} \\=\sqrt[3]{2^4\cdot2^4\cdot2^8}$ Use the rule $a^m\cdot a^n = a^{m+n}$ to obtain: $=\sqrt[3]{2^{4+4+8}} \\=\sqrt[3]{2^{16}}$ Factor the radicand so that one factor is a perfect cube to obtain: $=\sqrt[3]{(2^{15})(2)} \\=\sqrt[3]{(2^5)^3(2)}$ Bring out the cube root of the perfect cube factor to obtain: $=2^5\sqrt[3]{2} \\=\color{blue}{32\sqrt[3]{2}}$
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