Easy Algebric equation

the sum of all possible values of a root of 1 is always $\boxed{0}$ in this case, the roots are 1, $\frac{-1}{2} + \frac{\sqrt{3}i}{2}$ and $\frac{-1}{2} - \frac{\sqrt{3}i}{2}$

$\color{#3D99F6}{x=1^{\frac{1}{3}}\rightarrow x^3=1\rightarrow x^3-1=0}$ From Vieta,s formulas,we know that $r_1+r_2+r_3=0\;where\;r_1,r_2,r_3=Roots$ so answer is $\boxed{0}$

x = 1^(1/3)

cubing them,

x^3 = 1

x^3 - 1 = 0

(x-1)(x^2 + x + 1) = 0

x = 1

And,

x^2 + x + 1 = 0

x^2 + x + 1/4 = 1/4 - 1

(x+1/2)^2 = -3/4

x = (-1 +/- i√3)/2

Then, their sum is just, 1 + -1 or 0.