Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises: 67

Answer

$\color{blue}{\dfrac{2\sqrt{x^2}}{x^2}}$

Work Step by Step

Rationalize the denominator by multiplying $x^2$ to both the numerator and denominator of the radicand to obtain: $=\sqrt[3]{\dfrac{8(x^2)}{x^4(x^2)}} \\=\sqrt[3]{\dfrac{8x^2}{x^6}} \\=\sqrt[3]{\dfrac{8x^2}{(x^2)^3}}$ Bring out the cube root of the denominator to obtain: $\\=\dfrac{\sqrt[3]{8x^2}}{x^2}$ Factor the radicand such that at least one factor is a perfect cube to obtain: $\\=\dfrac{\sqrt[3]{(8(x^2)}}{x^2} \\=\dfrac{\sqrt{(2)^3(x^2)}}{x^2}$ Bring out the cube root of the perfect cube factor/s of the numerator to obtain: $\\=\color{blue}{\dfrac{2\sqrt{x^2}}{x^2}}$
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