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: 3



Work Step by Step

$64x^{6}$ $=(8)(8)(x)(x)(x)(x)(x)(x)$ $=(8)(8)(x^{3})(x^{3})$ $=(8x^{3})^{2}$
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