Answer
$(\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}i)^{3}=i$
So,$\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}i$ is a cube root of $i$.
Work Step by Step
$(\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}i)^{3}\qquad$...Combine the fractions
$(\displaystyle \frac{\sqrt{3}+i}{2})^{3}\qquad$...Apply exponent rule: $(\displaystyle \frac{a}{b})^{n}=\frac{a^{n}}{b^{n}}$
$=\displaystyle \frac{(\sqrt{3}+i)^{3}}{2^{3}}\qquad$...Apply Perfect Cube Formula: $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3},\ a=\sqrt{3},\ b=i$.
$=\displaystyle \frac{(\sqrt{3})^{3}+3(\sqrt{3})^{2}i+3\sqrt{3}i^{2}+i^{3}}{2^{3}}\qquad$...Simplify.
$=\displaystyle \frac{3\sqrt{3}+9i-3\sqrt{3}+i^{3}}{8}\qquad$...Group like terms
$=\displaystyle \frac{i^{3}+9i+3\sqrt{3}-3\sqrt{3}}{8}\qquad$...Simplify.
$=\displaystyle \frac{i^{3}+9i}{8}\qquad$...Simplify. $i^{3}=i^{2}\cdot i=-1\cdot i=-i$
$=\displaystyle \frac{-i+9i}{8}$
$=\displaystyle \frac{8i}{8}$
$=i$
Therefore, $(\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}i)^{3}=i$