Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapters 1-9 Cumulative Review Problem Set: 23

Answer

$3x^{3}y^{6}-4y^{3}$

Work Step by Step

We factor $5x^{3}y^{2}$ from the numerator. Since $5x^{3}y^{2}$ is now common to both the numerator and the denominator, it is canceled and the resulting expression simplified. $\frac{15x^{6}y^{8}-20x^{3}y^{5}}{5x^{3}y^{2}}$ =$\frac{5x^{3}y^{2}(3x^{3}y^{6}-4y^{3})}{5x^{3}y^{2}}$ =$\frac{1(3x^{3}y^{6}-4y^{3})}{1}$ =$(3x^{3}y^{6}-4y^{3})$ =$3x^{3}y^{6}-4y^{3}$
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