College Algebra (10th Edition)

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

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 48

Answer

$x=2$ or $x=-1$

Work Step by Step

We solve by factoring: $x^{6}-7x^{3}-8=0$ $(x^{3}-8)(x^{3}+1)=0$ $x^{3}-8=0$ or $x^{3}+1=0$ $x^{3}=8$ or $x^{3}=-1$ $x=\sqrt[3]{8}$ or $x=\sqrt[3]{-1}$ $x=2$ or $x=-1$
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