Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.4 - Factoring Special Forms - Exercise Set: 87

Answer

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

Work Step by Step

The given binomial can be written as: $(xy)^3-4^3$ The binomial above is a difference of two cubes. RECALL: A difference of two cubes can be factored using the formula: $a^3-b^3=(a-b)(a^2+ab+b^2)$ Use the formula above to factor the sum of two cubes with $a=xy$ and $b=4$ to obtain: $=(xy-4)[(xy)^2+xy(4) + 4^2] \\=(xy-4)(x^2y^2+4xy+16)$
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