College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.5 - Rational Expressions - R.5 Exercises - Page 47: 22

Answer

$y^2+3y+9$ $y\ne3$

Work Step by Step

Given. $\frac{y^3-27}{y-3}$ Write 27 as $3^3$. $=\frac{y^3-3^3}{y-3}$ Factor the sum of two cubes using the formula $a^3-b^3=(a-b)(a+ab+b^2)$. $=\frac{(y-3)(y^2+3y+9)}{y-3}$ Cancel the common factor $y-3$. $y^2+3y+9,y\ne3$
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