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 - Exercise Set: 89a



Work Step by Step

Using $a^2-b^2=(a+b)(a-b)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)$, then the given expression, $ x^6-1 $, is equivalent to \begin{array}{l} (x^3-1)(x^3+1) \\\\ (x-1)(x^2+x+1)(x^3+1) \\\\ (x-1)(x^2+x+1)(x+1)(x^2-x+1) .\end{array}
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