Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.5 Rational Expressions - R.5 Exercises - Page 54: 78

Answer

$\frac{x+y}{x^2+xy+y^2}$

Work Step by Step

Step 1. Factor the denominators: $x^3-y^3=(x-y)(x^2+xy+y^2)$ and $x^2-y^2=(x+y)(x-y)$ Step 2. Use the above results in the original expression, we have: $\frac{\frac{1}{x^3-y^3}}{\frac{1}{x^2-y^2}}=\frac{\frac{1}{(x-y)(x^2+xy+y^2)}}{\frac{1}{(x+y)(x-y)}}=\frac{(x+y)(x-y)}{(x-y)(x^2+xy+y^2)}=\frac{x+y}{x^2+xy+y^2}$
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