How do I use DeMoivre’s theorem to solve ##z^3-1=0##?

If ##z^3-1=0##, then we are looking for the cubic roots of unity, i.e. the numbers such that ##z^3=1##.
If you’re using complex numbers, then every polynomial equation of degree ##k## yields exactly ##k## solution. So, we’re expecting to find three cubic roots.
De Moivre’s theorem uses the fact that we can write any complex number as ##rho e^{i theta}= rho (cos(theta)+isin(theta))##, and it states that, if
##z=rho (cos(theta)+isin(theta))##, then
##z^n = rho^n (cos(n theta)+isin(n theta))##
If you look at ##1## as a complex number, then you have ##rho=1##, and ##theta=2pi##. We are thus looking for three numbers such that ##rho^3=1##, and ##3theta=2pi##.
Since ##rho## is a real number, the only solution to ##rho^3=1## is ##rho=1##. On the other hand, using the periodicity of the angles, we have that the three solutions for ##theta## are
##theta_{1,2,3}=frac{2kpi}{3}##, for ##k=0,1,2##.
This means that the three solutions are:
##rho=1, theta=0##, which is the real number ##1##.
##rho=1, theta=frac{2pi}{3}##, which is the complex number ##-1/2 + sqrt{3}/2 i##
##rho=1, theta=frac{4pi}{3}##, which is the complex number ##-1/2 – sqrt{3}/2 i##