## Precalculus (6th Edition)

$\color{blue}{3\sqrt[3]{4}}$
Simplify each radical. Factor the radicand so that one factor is a perfect cube, then bring out the cube root of the perfect cube factor to obtain: $=\sqrt[3]{8(4)}-5\sqrt[3]{4} +2\sqrt[3]{27(4)} \\=\sqrt[3]{2^3(3)} -5 \sqrt[3]{4}+2\sqrt[3]{3^3(4)} \\=2\sqrt[3]{4}-5\sqrt[3]{4}+2(3)\sqrt[3]{3} \\=2\sqrt[3]{4}-5\sqrt[3]{4}+6\sqrt[3]{3}$ Combine like terms to obtain: $=(2-5+6)\sqrt[3]{4} \\=\color{blue}{3\sqrt[3]{4}}$