Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises - Page 76: 110

Answer

$(\sqrt[3] {3^2}+\sqrt[3] 3 \sqrt[3] 5+\sqrt[3] {5^2})$ $-\frac{\sqrt[3] {3^2}+\sqrt[3] 3 \sqrt[3] 5+\sqrt[3] {5^2}}{2}$

Work Step by Step

The number should be $(\sqrt[3] {3^2}+\sqrt[3] 3 \sqrt[3] 5+\sqrt[3] {5^2})$. Multiply $\frac{\sqrt[3] {3^2}+\sqrt[3] 3 \sqrt[3] 5+\sqrt[3] {5^2}}{\sqrt[3] {3^2}+\sqrt[3] 3 \sqrt[3] 5+\sqrt[3] {5^2}}$ to the expression, we have: $\frac{1}{\sqrt[3] 3-\sqrt[3] 5}\cdot \frac{\sqrt[3] {3^2}+\sqrt[3] 3 \sqrt[3] 5+\sqrt[3] {5^2}}{\sqrt[3] {3^2}+\sqrt[3] 3 \sqrt[3] 5+\sqrt[3] {5^2}}=\frac{\sqrt[3] {3^2}+\sqrt[3] 3 \sqrt[3] 5+\sqrt[3] {5^2}}{3-5}=-\frac{\sqrt[3] {3^2}+\sqrt[3] 3 \sqrt[3] 5+\sqrt[3] {5^2}}{2}$
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