## Precalculus (6th Edition)

$\color{blue}{(3x-2x^2)\sqrt[4]{x^2y^3}}$
Simplify each radical. Factor the radicand so that one factor is a perfect fourth power, then bring out the fourth root of the perfect fourth power factor to obtain: $=\sqrt[4]{81x^4(x^2y^3)}-\sqrt[4]{16x^8(x^2y^3)} \\=\sqrt[4]{3^4x^4(x^2y^3)} - \sqrt[4]{2^4(x^2)^4(x^2y^3)} \\=3x\sqrt[4]{x^2y^3}-2x^2\sqrt[4]{x^2y^3}$ Combine like terms to obtain: $\\=\color{blue}{(3x-2x^2)\sqrt[4]{x^2y^3}}$
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