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 - Page 310: 90a

Answer

$(x-1)(x+1)(x^2+1)(x^2+x+1)(x^2-x+1)(x^4-x^2+1)$

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^{12}-1 $, is equivalent to \begin{array}{l} (x^3)^{4}-1^4 \\\\= [(x^3)^2-1^2][(x^3)^2+1^2] \\\\= (x^6-1)(x^6+1) \\\\= (x^3-1)(x^3+1)(x^6+1) \\\\= (x-1)(x^2+x+1)(x^3+1)(x^6+1) \\\\= (x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^6+1) \\\\= (x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1) \\\\= (x-1)(x+1)(x^2+1)(x^2+x+1)(x^2-x+1)(x^4-x^2+1) .\end{array}
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