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: 50



Work Step by Step

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of the difference of 2 cubes, then, \begin{array}{l} 125x^3+8y^3 \\= [(5x)+(2y)][(5x)^2-(5x)(2y)+(2y)^2] \\= (5x+2y)(25x^2-10xy+4y^2) .\end{array}
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