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: 66

Answer

$x^{2}y^{3}$

Work Step by Step

$\sqrt[4] (x^{8}y^{12})=x^{2}y^{3}$, because $(x^{2}y^{3})^{4}=x^{2}y^{3}\times x^{2}y^{3}\times x^{2}y^{3}\times x^{2}y^{3}=x^{2+2+2+2}\times y^{3+3+3+3}=x^{8}y^{12}$
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