Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.6 - Radical Equations and Problem Solving - Exercise Set - Page 453: 13

Answer

$x=29$

Work Step by Step

$\sqrt[3] (x-2)-3=0$ Add 3 to both sides of the equation. $\sqrt[3] (x-2)=3$ Cube both sides. $(\sqrt[3] (x-2))^{3}=3^{3}$ $x-2=27$ Add 2 to both sides. $x=29$
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