Discussion
The value of \(\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)^{3}\) + \(\left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)^{3}\) is equal to
Answer: D
The value is 0.
Write the numbers in polar form: \(\frac{\sqrt{3}}{2} \pm \frac{i}{2} = \cos 30^\circ \pm i\sin 30^\circ\). By De Moivre's theorem, cubing triples the angle.
\[ \begin{aligned} (\cos 30^\circ + i\sin 30^\circ)^{3} &= \cos 90^\circ + i\sin 90^\circ = i \\ (\cos 30^\circ - i\sin 30^\circ)^{3} &= \cos 90^\circ - i\sin 90^\circ = -i \end{aligned} \]
The sum is \(i + (-i) = 0\). The two terms are conjugates, so their sum is twice the real part of either, \(2\cos 90^\circ = 0\). \(2\sqrt{3}\) and \(3\sqrt{3}\) come from expanding the cubes with a sign error.
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