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: 35



Work Step by Step

Factoring the $GCF= a $ results to $ a(8b^3+27a^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} a(8b^3+27a^3) \\= a^2[(2b)+(3a)][(2b)^2-(2b)(3a)+(3a)^2) \\= a^2(2b+3a)(4b^2-6ab+9a^2) .\end{array}
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