College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Review Exercises - Page 145: 11


$x=\pm 3$

Work Step by Step

We solve: $\sqrt[3]{x^{2}-1}=2$ $(\sqrt[3]{x^{2}-1})^{3}=2^3$ $x^{2}-1=8$ $x^{2}=8+1$ $x^{2}=9$ $x=\pm \sqrt{9}$ $x=\pm 3$
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