# How do you use the binomial theorem to expand (-1/2+sqrt3/2i)^3?

Jul 7, 2018

The answer $= 1$

#### Explanation:

Normally, you don't need the binomial theorem but Euler's Identity and Demoivre's Theorem

${i}^{2} = - 1$

So, by the binomial theorem

${\left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)}^{3} = {\left(- \frac{1}{2}\right)}^{3} + 3 {\left(- \frac{1}{2}\right)}^{2} \left(\frac{\sqrt{3}}{2} i\right) + 3 \left(- \frac{1}{2}\right) {\left(\frac{\sqrt{3}}{2} i\right)}^{2} + {\left(\frac{\sqrt{3}}{2} i\right)}^{3}$

$= - \frac{1}{8} + \frac{3 \sqrt{3}}{8} i + \frac{9}{8} - \frac{3 \sqrt{3}}{3} i$

$= 1$