What is ( 3 +i 3 -i )^3 equal to?

Rationalizing the expression inside the bracket:

$\dfrac{\sqrt{3}+i}{\sqrt{3}-i} = \dfrac{(\sqrt{3}+i)^2}{(\sqrt{3}-i)(\sqrt{3}+i)}$

$= \dfrac{3 + i^2 + 2\sqrt{3}i}{3 - i^2}$

$= \dfrac{3 - 1 + 2\sqrt{3}i}{3 - (-1)}$

$= \dfrac{2 + 2\sqrt{3}i}{4} = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$

Converting to polar form:

\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i = \cos \left(\dfrac{\pi}{3}\right) +i\sin \left(\dfrac{\pi}{3}\right)<br><br>Using De Moivre's theorem to find the cube:<br><br>\left(\cos\left(\dfrac{\pi}{3}\right) + i\\sin \left(\dfrac{\pi}{3}\right)\right)^ 3 = \cos\left(3 \times $\dfrac{\pi}{3}$\right) + i\sin\left(3 \times \dfrac{\pi}{3}\right)

$= \cos\pi +i\sin\pi$

$= -1 + 0 = -1$

Answer: $-1