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: 63

Answer

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

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, $ 64t^6-1 $, is \begin{array}{l} (8t^3+1)(8t^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, $ (8t^3+1)(8t^3-1) $ is \begin{array}{l} (2t+1)[(2t)^2-(2t)(1)+(1)^2](2t-1)[(2t)^2-(2t)(-1)+(-1)^2] \\\\= (2t+1)(4t^2-2t+1)(2t-1)(4t^2+2t+1) \\\\= (2t+1)(2t-1)(4t^2-2t+1)(4t^2+2t+1) .\end{array}
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