Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 54



Work Step by Step

Factoring the $GCF= x^2y^3 $ results to $ x^2y^3(y^6+1) $. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of the sum of 2 cubes, then, \begin{array}{l} x^2y^3(y^6+1) \\= x^2y^3[(y^2)+(1)][(y^2)^2-(y^2)(1)+(1)^2] \\= x^2y^3(y^2+1)(y^4-y^2+1) .\end{array}
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