Intermediate Algebra (6th Edition)

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

Chapter 6 - Section 6.1 - Rational Functions and Multiplying and Dividing Rational Expressions - Exercise Set: 62

Answer

$\dfrac{m^{3}-n^{3}}{m-n}=m^{2}+mn+n^{2}$

Work Step by Step

$\dfrac{m^{3}-n^{3}}{m-n}$ The numerator is a difference of cubes. Factor it: $\dfrac{m^{3}-n^{3}}{m-n}=\dfrac{(m-n)(m^{2}+mn+n^{2})}{m-n}=...$ Simplify by removing the factors that appear both in the numerator and in the denominator of the resulting expression: $...=m^{2}+mn+n^{2}$
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.