<em>x</em> ³ = 1
<em>x</em> ³ - 1 = 0
<em>x</em> ³ - 1³ = 0
Factorize the right side as a difference of cubes:
(<em>x</em> - 1) (<em>x</em> ² + <em>x</em> + 1) = 0
Then
<em>x</em> - 1 = 0 <u>or</u> <em>x</em> ² + <em>x</em> + 1 = 0
The first equation yields <em>x</em> = 1 as a root.
For the other equation, rearrange and complete the square to get the other two roots,
<em>x</em> ² + <em>x</em> + 1/4 = -3/4
(<em>x</em> + 1/2)² = -3/4
<em>x</em> + 1/2 = ±√(-3/4) = ± <em>i</em> √3/2
<em>x</em> = -1/2 ± <em>i</em> √3/2
<em>x</em> = (-1 ± <em>i</em> √3)/2
Since complex roots occurs in conjugate pairs, taking the sum of these two roots eliminates the imaginary part:
(-1 + <em>i</em> √3)/2 + (-1 - <em>i</em> √3)/2 = (-1 - 1)/2 = -2/2 = -1
and adding to 1 indeed shows that the sum of roots is zero.