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: 78

Answer

$9\sqrt[3]{3}$

Work Step by Step

Simplify each radicand to obtain $3\sqrt[3]{8(3)}+\sqrt[3]{27(3)} \\=3\sqrt[3]{2^3(3)}+\sqrt[3]{3^3(3)} \\=3\cdot2\sqrt[3]{3} + 3\sqrt [3]{3} \\=6\sqrt[3]{3} + 3\sqrt[3]{3}.$ RECALL: The distributive property states that for any real numbers a, b, and c, $ac + bc = (a+b)c.$ Use the distributive to obtain $(6+3)\sqrt[3]{3} \\=9\sqrt[3]{3}.$
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