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: 56



Work Step by Step

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