### Instruction: Compute $(2 \cos 20^\circ + 2i \sin 20^\circ)^6.$  Enter your answer in rectangular form.### Response: I recognize that this expression is the complex number $2 cis 20^\circ$, where $cis x = \cos x + i \sin x$.
I also know that multiplying complex numbers in polar form is equivalent to adding their arguments and multiplying their magnitudes.
So, raising $2 cis 20^\circ$ to the sixth power is the same as multiplying its argument by 6 and its magnitude by itself 6 times.
That means $(2 cis 20^\circ)^6 = 2^6 cis 120^\circ = 64 cis 120^\circ$.
To convert this back to rectangular form, I use the definition of $cis x$ and get $64 (\cos 120^\circ + i \sin 120^\circ)$.
I recall that $\cos 120^\circ = -1/2$ and $\sin 120^\circ = \sqrt{3}/2$, so I can simplify the expression further.