## Precalculus (6th Edition)

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