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: 59

Answer

$5\left(xy^2-\dfrac{1}{2}\right)\left(x^2y^4+\dfrac{xy^2}{2}+\dfrac{1}{4}\right)$

Work Step by Step

Factoring the $GCF= 5 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 5x^3y^6-\dfrac{5}{8} \\\\= 5\left(x^3y^6-\dfrac{1}{8}\right) .\end{array} The expressions $ x^3y^6 $ and $ \dfrac{1}{8} $ are both perfect cubes (the cube root is exact). Hence, $ x^3y^6-\dfrac{1}{8} $ is a $\text{ difference }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} 5\left[\left(xy^2\right)^3-\left(\dfrac{1}{2}\right)^3\right] \\\\= 5\left(xy^2-\dfrac{1}{2}\right)\left[(xy^2)^2+xy^2\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2 \right] \\\\= 5\left(xy^2-\dfrac{1}{2}\right)\left(x^2y^4+\dfrac{xy^2}{2}+\dfrac{1}{4}\right) .\end{array}
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