Chapter 9 - Polar Coordinates; Vectors - 9.3 The Complex Plane; De Moivre's Theorem - 9.3 Assess Your Understanding - Page 594: 47

$\frac{27}{2}+i\frac{27\sqrt3}{2}.$

Using De Moivre's Theorem ($(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}.$): $[\sqrt3(\cos{(10^\circ)}+i(\sin{(10^\circ)})]^6=\sqrt3^6(\cos{(6\cdot10^\circ)}+i(\sin{(6\cdot10^\circ)}))=27\cdot(\cos{(60^\circ)}+i(\sin{(60^\circ})))=27\cdot(\frac{1}{2}+i\frac{\sqrt3}{2})=\frac{27}{2}+i\frac{27\sqrt3}{2}.$