Solving $\cos x + \cos 2x - \cos 3x = 1$ with the substitution $z = \cos x + i \sin x$ I need to solve 
$$\cos x+\cos 2x-\cos 3x=1$$
using the substitution$$z= \cos x + i \sin x $$
I fiddled around with the first equation using the double angle formula and addition formula to get 
$$\cos^2 x+4 \sin^2x\cos x-\sin^2 x=1$$ 
which gets me pretty close to something into which I can substitute $z$, because $$z^2= \cos^2 x-\sin^2 x+2i\sin x\cos x$$ 
I have no idea where to go from there.

Using the hint,
$$\Re(z+z^2-z^3)=1,$$ or
$$z+z^2-z^3=1+iw.$$
This can be factored as
$$-(z+1)(z-1)^2=iw$$ but I see no easy way to exploit it.
Direct resolution of the cubic equation looks terrible.
