[tex]\sqrt3+i=2\left(\dfrac{\sqrt3}2+\dfrac12i\right)[/tex]
[tex]\begin{cases}\cos\dfrac\pi6=\dfrac{\sqrt3}2\\\\\sin\dfrac\pi6=\dfrac12\end{cases}[/tex]
[tex]\implies\sqrt3+i=2e^{i\pi/6}[/tex]
By DeMoivre's theorem,
[tex](\sqrt3+i)^3=(2e^{i\pi/6})^3=8e^{i\pi/2}=8\left(\cos\dfrac\pi2+i\sin\dfrac\pi2\right)=8i[/tex]
