Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 24

Answer

$(3x-4y)(9x^2+12xy+16y^2)$

Work Step by Step

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of two cubes, then, \begin{array}{l} 27x^3-64y^3 \\= [(3x)+(-4y)][(3x)^2-(3x)(-4y)+(-4y)^2] \\= (3x-4y)(9x^2+12xy+16y^2) .\end{array}
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