Solve
$x=1+w^x + w^{2x}$
Where $w$ ,$w^2$ are cube roots of the unit and $x$ is pure real number My attempt :-
let $w = \cos(120) + i \sin(120)$
Then
$x = 1+\cos(120x) + \cos(240x) + [\sin(120x) + \sin(240x) ]i$
$x = 1+\cos(120x) + \cos(240x) $
And
$\sin(120x) + \sin(240x) = 0$
I can solve the second equation i fo not know how can i solve the first one ?
$x=1+w^x+w^{2x}\implies x(w^x-1)=w^{3x}-1=0\implies x=0 $ or $w^x=1$.
But $x=0$ is not a solution, so $x=1+w^x+w^{2x}=1+1+1=3$.