College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3: 78

Answer

$9\sqrt[3]{3}$

Work Step by Step

$$3\sqrt[3]{24} + \sqrt[3]{81}$$ $$=3\sqrt[3]{8\times3} + \sqrt[3]{27\times3}$$ $$=(3\times2)\sqrt[3]{3} +3\sqrt[3]{3}$$ $$=6\sqrt[3]{3} + 3\sqrt[3]{3}$$ $$=(6+3)\sqrt[3]{3}$$ $$=9\sqrt[3]{3}$$
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