Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set: 71

Answer

$3\sqrt[3]{2}$

Work Step by Step

RECALL: For any real numbers $a$ and $b$, $\sqrt[3]{a} \cdot \sqrt[3]{b} = \sqrt[3]{ab}.$ Use the rule above to obtain $\sqrt[3]{9(6)} \\=\sqrt{54}.$ Factor the radicand so that one factor is a perfect cube to obtain $\sqrt[3]{27(2)} \\=\sqrt[3]{3^3(2)}.$ Simplify to obtain $3\sqrt[3]{2}.$
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