Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.6 - Radical Equations - Exercise Set - Page 559: 18

Answer

$x=5$

Work Step by Step

RECALL: For any real number $a$, $(\sqrt[3]{a})^3=a$ Cube both sides then use the rule above to simplify: $(\sqrt[3]{6x-3})^3 = 3^3 \\6x-3=27$ Add 3 on both sides to obtain: $6x = 27+3 \\6x=30$ Divide 6 on both sides to obtain: $x=5$
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