Answer
$(\sqrt{3} + i)^3$
= $8i$
Work Step by Step
$(\sqrt{3} + i)$ is at $30^\circ$ with absolute value $\sqrt{(\sqrt{3})^2 + 1^2}=2$
$(\sqrt{3} + i)^3$
= $(2cis30^\circ)^3$
= $2^3[cos(3\cdot 30^\circ) + isin(3\cdot 30^\circ)]$ (De Moivre’s Theorem)
= $8(cos90^\circ + isin90^\circ)$
= $8(0 + i)$
= $8i$
