## Precalculus (10th Edition)

$\frac{27}{2}\frac{-27\sqrt3}{2}i.$
Using De Moivre's Theorem ($(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}.$) and a calculator: $[\sqrt3(\cos{(\frac{5\pi}{18})}+i(\sin{(\frac{5\pi}{18})})]^6=\sqrt3^6(\cos{(6\cdot\frac{5\pi}{18})}+i(\sin{(6\cdot\frac{5\pi}{18})}))=27\cdot(\cos{(\frac{5\pi}{3})}+i(\sin{(\frac{5\pi}{3}})))=27\cdot(\frac{1}{2}+\frac{-\sqrt3}{2}i)=\frac{27}{2}\frac{-27\sqrt3}{2}i.$