Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-2 Multiplying and Dividing Radical Expressions - Practice and Problem-Solving Exercises - Page 371: 46



Work Step by Step

Multiplying the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index, the given expression, $ \sqrt[3]{\dfrac{5}{3x}} ,$ is equivalent to \begin{align*} & =\sqrt[3]{\dfrac{5}{3x}\cdot\dfrac{3^2x^2}{3^2x^2}} \\\\&= \sqrt[3]{\dfrac{45x^2}{3^3x^3}} \\\\&= \sqrt[3]{\dfrac{45x^2}{(3x)^3}} .\end{align*} Using $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}},$ the expression above is equivalent to \begin{align*} & =\dfrac{\sqrt[3]{45x^2}}{\sqrt[3]{(3x)^3}} \\\\&= \dfrac{\sqrt[3]{45x^2}}{3x} .\end{align*} Hence, the rationalized-denominator form of the given expression is $ \dfrac{\sqrt[3]{45x^2}}{3x} $.
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