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 310: 62

Answer

$(xy+6)(x^2y^2-6xy+36)$

Work Step by Step

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^3y^3+216 \\= [(xy)+(6)][(xy)^2-(xy)(6)+(6)^2] \\= (xy+6)(x^2y^2-6xy+36) .\end{array}
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