Chapter 8 - Polar Coordinates; Vectors - Section 8.3 The Complex Plane; De Moivre's Theorem - 8.3 Assess Your Understanding - Page 615: 45

$\dfrac{27}{2}+\dfrac{27\sqrt {3}}{2}i$

Applying De Moivre's theorem, we have \begin{align*} [\sqrt {3}(\cos 10^{\circ}+i\sin 10^{\circ})]^{6}&=(\sqrt {3})^{6}[\cos(6\cdot10^{\circ})+i\sin(6\cdot10^{\circ})]\\ &=27(\cos 60^{\circ}+i\sin 60^{\circ})\\ &=27\left(\frac{1}{2}+\frac{\sqrt {3}}{2}i\right)\\ &= \frac{27}{2}+\frac{27\sqrt {3}}{2}i\end{align*}