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 - 6-3 Binomial Radical Expressions - Practice and Problem-Solving Exercises - Page 380: 75

Answer

$\dfrac{2\sqrt[3]{x^2}}{x}$

Work Step by Step

Using $\sqrt[n]{\dfrac{x}{y}}=\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}},$ the given expression is equivalent to: \begin{align*} & =\dfrac{\sqrt[3]4}{\sqrt[3]{0.5x}} \end{align*} Multiplying the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index, the expression above is equivalent to \begin{align*} & =\dfrac{\sqrt[3]4}{\sqrt[3]{0.5x}}\cdot\dfrac{\sqrt[3]{0.5^2x^2}}{\sqrt[3]{0.5^2x^2}} \\\\&= \dfrac{\sqrt[3]{4(0.25x^2)}}{\sqrt[3]{0.5^3x^3}} \\\\&= \dfrac{\sqrt[3]{x^2}}{\sqrt[3]{(0.5x)^3}} \\\\&= \dfrac{\sqrt[3]{x^2}}{0.5x} \\\\&= \dfrac{\sqrt[3]{x^2}}{\frac{1}{2}x} \\\\&= \dfrac{2\sqrt[3]{x^2}}{x} \end{align*} Hence, the simplified form of the given expression is $ \dfrac{2\sqrt[3]{x^2}}{x} $.
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