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: 34


$\left( x-\dfrac{1}{2} \right)\left( x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)$

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-\dfrac{1}{8} ,$ is \begin{array}{l} \left( x-\dfrac{1}{2} \right)\left[ (x)^2-(x)\left( -\dfrac{1}{2} \right)+\left( -\dfrac{1}{2} \right)^2\right] \\\\= \left( x-\dfrac{1}{2} \right)\left( x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right) .\end{array}
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