### Instruction: Compute $(2 \cos 20^\circ + 2i \sin 20^\circ)^6.$  Enter your answer in rectangular form.### Response: Let's use De Moivre's Theorem.
Right. We can express $2\cos 20^\circ + 2i\sin 20^\circ$ in polar coordinates as $2\operatorname{cis} 20^\circ.$
Then according to De Moivre's Theorem, $(2\operatorname{cis} 20^\circ)^6 = 2^6 \operatorname{cis} (6 \cdot 20^\circ) = 64 \operatorname{cis} 120^\circ.$
And $\operatorname{cis} 120^\circ = -\frac{1}{2} + \frac{\sqrt{3}}{2} i.$