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 560: 28


$x = 11$

Work Step by Step

Subtract 5 on both sides to obtain: $(3x-6)^{\frac{1}{3}}=3$ Cube both sides to obtain: $\left((3x-6)^{\frac{1}{3}}\right)^3=3^3 \\(3x-6)^{\frac{1}{3}(3)}=27 \\(3x-6)^1=27 \\3x-6=27$ Aadd 6 on both sides to obtain: $3x=33$ Divide 3 on both sides to obtain: $x = 11$
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