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: 7

Answer

$\frac{4\sqrt[3] 9 }{3}$

Work Step by Step

$\frac{4}{\sqrt[3] 3}=\frac{4 \times \sqrt[3] (3^{2}) }{\sqrt[3] 3 \times \sqrt[3] (3^{2})}=\frac{4\times\sqrt[3] (3^{2}) }{\sqrt[3] (3\times3^{2})}=\frac{4\sqrt[3] (3^{2}) }{\sqrt[3] 27}=\frac{4\sqrt[3] 9 }{3}$ We know that $\sqrt[3] 27=3$, because $3^{3}=27$.
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