Find all three distinct roots for X³ =1. (two of them are complex mumber). And show that sum of all roots is zero.​

x ³ = 1

x ³ - 1 = 0

x ³ - 1³ = 0

Factorize the right side as a difference of cubes:

(x - 1) (x ² + x + 1) = 0

Then

x - 1 = 0   or   x ² + x + 1 = 0

The first equation yields x = 1 as a root.

For the other equation, rearrange and complete the square to get the other two roots,

x ² + x + 1/4 = -3/4

(x + 1/2)² = -3/4

x + 1/2 = ±√(-3/4) = ± i √3/2

x = -1/2 ± i √3/2

x = (-1 ± i √3)/2

Since complex roots occurs in conjugate pairs, taking the sum of these two roots eliminates the imaginary part:

(-1 + i √3)/2 + (-1 - i √3)/2 = (-1 - 1)/2 = -2/2 = -1

and adding to 1 indeed shows that the sum of roots is zero.