College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding - Page 71: 32


$-\displaystyle \frac{9x^{3}}{(x-3)^{2}}$

Work Step by Step

$...=\displaystyle \frac{3+x}{3-x}\div\frac{x^{2}-9}{9x^{3}}=\frac{3+x}{3-x}\cdot\frac{9x^{3}}{x^{2}-9}$ Factoring, $ x^{2}-9=(x+3)(x-3)\quad$ (a difference of squares) $3-x=-(x-3)$ ... = $\displaystyle \frac{(x+3)\cdot 9x^{3}}{-(x-3)(x-3)(x+3)}$ ... cancel the following from both sides of the fraction line: $(x+3)$ ... = $\displaystyle \frac{9x^{3}}{-(x-3)(x-3)}$ = $-\displaystyle \frac{9x^{3}}{(x-3)^{2}}$
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