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

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Answer 1 · 2018-07-07T15:44:27

The answer $=1$

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

$i^2=-1$

So, by the binomial theorem

$(-1/2+sqrt3/2i)^3=(-1/2)^3+3(-1/2)^2(sqrt3/2i)+3(-1/2)(sqrt3 /2i)^2+(sqrt3/2i)^3$

$=-1/8+(3sqrt3)/8i+9/8-(3sqrt3)/3i$

$=1$

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