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 53: 46

Answer

$x-y$

Work Step by Step

Factor the numerators and the denominators, then simplify, we have: $\frac{x^2-y^2}{(x-y)^2}\cdot\frac{x^2-xy+y^2}{x^2-2xy+y^2}\div\frac{x^3+y^3}{(x-y)^4}=\frac{(x+y)(x-y)}{(x-y)^2}\cdot\frac{x^2-xy+y^2}{(x-y)^2}\cdot\frac{(x-y)^4}{x^3+y^3}=\frac{(x+y)}{(x-y)}\cdot\frac{x^2-xy+y^2}{(x-y)^2}\cdot\frac{(x-y)^4}{(x+y)(x^2-xy+y^2)}=\frac{1}{1}\cdot\frac{1}{1}\cdot\frac{(x-y)}{1}=x-y$ where $x \pm y\ne0$ and $x^2-xy+y^2\ne 0$
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