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: 49



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} 8x^3+27y^3 \\= [(2x)+(3y)][(2x)^2-(2x)(3y)+(3y)^2] \\= (2x+3y)(4x^2-6xy+9y^2) .\end{array}
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