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: 32

Answer

$\color{blue}{y^2+3y+9}$

Work Step by Step

The given expression is equivalent to: $=\dfrac{y^3-3^3}{y-3}$ Factor the numerator and the denominator then cancel the common factors. Factor the numerator using the formula $a^3+-b^3=(a-b)(a^2+ab+b^2)$ with $a=y$ and $b=3$. : $\require{cancel} =\dfrac{(y-3)(y^2+3y+9)}{y-3} \\=\dfrac{\cancel{(y-3)}(y^2+3y+9)}{\cancel{y-3}} \\=\color{blue}{y^2+3y+9}$
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