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: 33


$\left( y-\dfrac{1}{10} \right) \left( y^2+\dfrac{1}{10}y+\dfrac{1}{100}\right)$

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, $ y^3-\dfrac{1}{1000} ,$ is \begin{array}{l} \left( y-\dfrac{1}{10} \right)\left[ (y)^2-(y)\left( -\dfrac{1}{10} \right)+\left( -\dfrac{1}{10} \right)^2\right] \\\\= \left( y-\dfrac{1}{10} \right) \left( y^2+\dfrac{1}{10}y+\dfrac{1}{100}\right) .\end{array}
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