Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Test - Page 402: 28

Answer

$(x-y),(x+y),(x^2+y^2),(x^4+y^4)$

Work Step by Step

Given: $x^8-y^8$ Take out the common terms as follows: $=(x^4)^2-(y^4)^2$ or, $=(x^4-y^4)(x^4+y^4)$ Thus, we have $=(x^2-y^2)(x^2+y^2)(x^4+y^4)$ or, $=(x-y)(x+y)(x^2+y^2)(x^4+y^4)$ factors are $(x-y),(x+y),(x^2+y^2),(x^4+y^4)$
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