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: 71

Answer

$3\sqrt[3] 2$

Work Step by Step

$\sqrt[3] 9\times \sqrt[3] 6=\sqrt[3] (9\times6)=\sqrt[3] 54=\sqrt[3] (27\times2)=\sqrt[3] 27\times \sqrt[3] 2=3\sqrt[3] 2$ We know that $\sqrt[3] 27=3$, because $3^{3}=27$.
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