Complex cube root of unity
\[x^3-1=0\]
Step1. Using hit and trial method x=1 is one of the roots of the equation.
\[x^3-1=(x-1)(x^2+x+1)\]
\[(x-1)(x^2+x+1)=0\]
\[(x^2+x+1)=0\]
let \[ \alpha \] and \[ \beta \] are the roots of the equation \[ x^2+x+1\]=0
where \[ \alpha \] and \[ \beta \] can be found using Sridhar Acharya formula
\[\alpha =\frac{-1+\sqrt{3}i}{2}\]
\[\beta =\frac{-1-\sqrt{3}i}{2}\]
Properties of complex numbers
let \[\omega\]=\frac{-1+\sqrt{3}i}{2}\]
and \[\omega ^2=\frac{-1-\sqrt{3}i}{2}\]
\[\omega^3=1\]
\[ 1+\omega +\omega ^2=0\]
