Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.1 - Radicals and Radical Functions - Exercise Set - Page 416: 10

Answer

$x^{8}$

Work Step by Step

$\sqrt (x^{16})=x^{8}$, because $(x^{8})^{2}=x^{8}\times x^{8}=x^{16}$
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