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: 36



Work Step by Step

Factoring the $GCF= b $ results to $ b(a^3+8b^3) $. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$, or the factoring of the sum/difference of two cubes, then, \begin{array}{l} b(a^3+8b^3) \\= b[(a)+(2b)][(a)^2-(a)(2b)+(2b)^2) \\= b(a+2b)(a^2-2ab+4b^2) .\end{array}
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