Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.7 - Factoring by Special Products - Vocabulary, Readiness & Video Check - Page 309: 8

Answer

$(xy^{2})^{3}$

Work Step by Step

$x^{3}y^{6}$ $=(x)(x)(x)(y)(y)(y)(y)(y)(y)$ $=(x)(x)(x)(y^{2})(y^{2})(y^{2})$ $=(xy^{2})^{3}$
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