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: 23

Answer

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

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, $ a^3+\dfrac{1}{8} ,$ is \begin{array}{l} \left( a+\dfrac{1}{2} \right)\left[ (a)^2-(a)\left( \dfrac{1}{2} \right)+\left( \dfrac{1}{2} \right)^2 \right] \\\\= \left( a+\dfrac{1}{2} \right)\left( a^2-\dfrac{1}{2}a+\dfrac{1}{4} \right) .\end{array}
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