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: 30



Work Step by Step

Using $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$, or the factoring of the sum/difference of two cubes, then, \begin{array}{l} m^3+n^3 \\= (m+n)[(m)^2-(m)(n)+(n)^2) \\= (m+n)(m^2-mn+n^2) .\end{array}$
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