## Precalculus (6th Edition)

$(-\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}i)^{3}=i$ So,$-\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}i$ is a cube root of $i$.
$(-\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}i)^{3}=i\qquad$...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. $=\displaystyle \frac{-3\sqrt{3}+9i+3\sqrt{3}-i}{8}\qquad$...Group like terms $=\displaystyle \frac{9i-i-3\sqrt{3}+3\sqrt{3}}{8}\qquad$...Simplify. $=\displaystyle \frac{8i}{8}$ $=i$ Therefore, $(-\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}i)^{3}=i$ So,$-\displaystyle \frac{\sqrt{3}}{2}+\frac{1}{2}i$ is a cube root of $i$.