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 - Page 338: 35

Answer

$(x-0.1)(x^2+0.1x+0.01)$

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, $ x^3-0.001 ,$ is \begin{array}{l} (x-0.1)[ (x)^2-(x)(-0.1)+(-0.1)^2] \\\\= (x-0.1)(x^2+0.1x+0.01) .\end{array}
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