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 - Vocabulary, Readiness & Video Check: 11

Answer

$4y^{3}$

Work Step by Step

$(4y^{3){2}} = 4^{2} \times y^{6} = 16y^{6}$ or... $\sqrt 16 = 4, and \sqrt y^{6} = y^{3} = 4y^{3}$
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