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: 42

Answer

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

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, $ t^6+1 ,$ is \begin{array}{l} (t^2+1)[ (t^2)^2-(t^2)(1)+(1)^2] \\\\= (t^2+1)(t^4-t^2+1) .\end{array}
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