Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 9 - Solving Quadratic Equations - 9.1 - Properties of Radicals - Exercises - Page 487: 90

Answer

$-\sqrt[3]{10}$

Work Step by Step

The given expression is $=\sqrt[3]{2}(\sqrt[3]{135}-4\sqrt[3]{5})$ Factor as cube term. $=\sqrt[3]{2}(\sqrt[3]{27\cdot 5}-4\sqrt[3]{5})$ Use product property of cube roots. $=\sqrt[3]{2}(\sqrt[3]{27}\cdot \sqrt[3]{5}-4\sqrt[3]{5})$ Simplify. $=\sqrt[3]{2}(3 \sqrt[3]{5}-4\sqrt[3]{5})$ Factor out $\sqrt[3]{5}$. $=\sqrt[3]{2}\sqrt[3]{5}(3-4)$ Simplify. $=\sqrt[3]{2}\sqrt[3]{5}(-1)$ Use product property of cube roots. $=-\sqrt[3]{2\cdot 5}$ $=-\sqrt[3]{10}$
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