Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.4 - Adding, Subtracting, and Dividing Radical Expressions - Exercise Set - Page 539: 62

Answer

$2x^2y^2\sqrt[5] {y^3}$.

Work Step by Step

The given expression is $\frac{\sqrt[5] {96x^{12}y^{11}}}{\sqrt[5] {3x^2y^{-2}}}$ Use the quotient rule $\frac{\sqrt[n] a}{\sqrt[n] b} = \sqrt[n] {\frac{a}{b}}$. $=\sqrt[5] \frac{{96x^{12}y^{11}}}{ {3x^2y^{-2}}}$ Use $\frac{a^m}{a^n}=a^{m-n}$ $=\sqrt[5] {32x^{12-2}y^{11-(-2)}}$ $=\sqrt[5] {32x^{12-2}y^{11+2}}$ Simplify. $=\sqrt[5] {32x^{10}y^{13}}$ Factor the radicand: $=\sqrt[5] {2^5x^{5}x^5y^{5}y^5y^3}$ Simplify. $=2xxyy\sqrt[5] {y^3}$ $=2x^2y^2\sqrt[5] {y^3}$.
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