Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set: 8

Answer

$2\sqrt[3] 3$

Work Step by Step

$\frac{6}{\sqrt[3] 9}=\frac{6 \times \sqrt[3] (9^{2}) }{\sqrt[3] 9 \times \sqrt[3] (9^{2})}=\frac{6\times\sqrt[3] (9^{2}) }{\sqrt[3] (9\times9^{2})}=\frac{6\sqrt[3] (9^{2}) }{\sqrt[3] (9^{3})}$ $=\frac{6\sqrt[3] 81}{9}=\frac{6\sqrt[3] (27\times3)}{9}=\frac{(6\times3)\sqrt[3] 3}{9}=\frac{18\sqrt[3] 3}{9}=2\sqrt[3] 3$ We know that $\sqrt[3] 27=3$, because $3^{3}=27$.
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