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 339: 57


$\dfrac{1}{2}\left(\dfrac{1}{2}x^{a}+y^{2a}z^{3b} \right)\left(\dfrac{1}{4}x^{2a}-\dfrac{1}{2}x^{a}y^{2a}z^{3b}+y^{4a}z^{6b} \right)$

Work Step by Step

Factoring the $GCF= \dfrac{1}{2} ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{16}x^{3a}+\dfrac{1}{2}y^{6a}z^{9b} \\\\= \dfrac{1}{2}\left(\dfrac{1}{8}x^{3a}+y^{6a}z^{9b} \right) .\end{array} The expressions $ \dfrac{1}{8}x^{3a} $ and $ y^{6a}z^{9b} $ are both perfect cubes (the cube root is exact). Hence, $ \dfrac{1}{8}x^{3a}+y^{6a}z^{9b} $ is a $\text{ sum }$ 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} \dfrac{1}{2}\left[\left(\dfrac{1}{2}x^{a}\right)^3+\left(y^{2a}z^{3b} \right)^3\right] \\\\= \dfrac{1}{2}\left(\dfrac{1}{2}x^{a}+y^{2a}z^{3b} \right)\left(\dfrac{1}{4}x^{2a}-\dfrac{1}{2}x^{a}y^{2a}z^{3b}+y^{4a}z^{6b} \right) .\end{array}
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