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: 55



Work Step by Step

The expressions $ x^{6a} $ and $ y^{3b} $ are both perfect cubes (the cube root is exact). Hence, $ x^{6a}-y^{3b} $ 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} (x^{2a})^3-(y^{b})^3 \\\\= (x^{2a}-y^{b})[(x^{2a})^2+x^{2a}(y^{b})+(y^{b})^2] \\\\= (x^{2a}-y^{b})(x^{4a}+x^{2a}y^{b}+y^{2b}) .\end{array}
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