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 - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set: 41

Answer

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

Work Step by Step

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $ z^6-1 ,$ is \begin{array}{l} (z^2-1)[ (z^2)^2-(z^2)(-1)+(-1)^2] \\\\= (z^2-1)(z^4+z^2+1) .\end{array} Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, then the complete factored form of the expression above is \begin{array}{l} (z+1)(z-1)(z^4+z^2+1) .\end{array}
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