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: 87

Answer

$\color{blue}{10}$

Work Step by Step

RECALL: $a^3-b^3=(a-b)(a^2+ab+b^2)$ Use the formula above with $a=\sqrt[3]{11}$ and $b=1$ to obtain: $(\sqrt[3]{11}-1)(\sqrt{11^2}+\sqrt[3]{11}+1) \\=(\sqrt[3]{11})^3-1^3 \\=11-1 \\=\color{blue}{10}$
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