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: 11

Answer

$4y^{3}$

Work Step by Step

$\sqrt (16y^{6})=4y^{3}$, because $(4y^{3})^{2}=4y^{3}\times 4y^{3}=(4\times4)\times(y^{3}\times y^{3})=16y^{6}$
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