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: 88

Answer

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

Work Step by Step

The given binomial can be written as: $(xy)^3-3^3$ 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 difference of two cubes with $a=xy$ and $b=3$ to obtain: $=(xy-3)[(xy)^2+xy(3) + 3^2] \\=(xy-3)(x^2y^2+3xy+9)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.