Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.6 Factoring: A General Strategy - 5.6 Exercise Set: 64

Answer

$(m+1)(m-1)(m^2-m+1)(m^2+m+1)$

Work Step by Step

Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, then the factored form of the given expression, $ m^6-1 $, is \begin{array}{l} (m^3+1)(m^3-1) .\end{array} Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of $2$ cubes, then the factored form of the expression, $ (m^3+1)(m^3-1) $ is \begin{array}{l} (m+1)[(m)^2-(m)(1)+(1)^2](m-1)[(m)^2-(m)(-1)+(-1)^2] \\\\= (m+1)(m^2-m+1)(m-1)(m^2+m+1) \\\\= (m+1)(m-1)(m^2-m+1)(m^2+m+1) .\end{array}
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