Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Review Exercises - Page 400: 84



Work Step by Step

Given: $27x^3-125y^3$ This implies that $=(3x)^3-(5y)^3$ or, $=(3x-5y)[(3x)^2+(5y)^2+(3x)(5y)]$ Hence, our Factors are: $(3x-5y),(9x^2+25y^2+15xy)$
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