Question

if the cube root of unity are 1, w;w2,then the roots of the equation x-1 whole cube +8=0

Answer 1

Since 1, ω, and ωâ‚‚ are the cube roots of unity (the three cube roots of 1), they are the three solutions to x^3 = 1 (note: ω and ωâ‚‚ are the two complex solutions to this).
If we let u = x - 1, then the equation becomes:
u³ + 8 = (u + 2)(u² - 2u + 4) = 0.
(note: u³ + 8 was factored via difference of cubes.)
So, the solutions occur when u = -2 (giving -2 = x - 1 => x = -1), or when:
u² - 2u + 4 = 0,

which has roots, by the Quadratic Formula, to be u = 1 ± i√3, and so:
x - 1 = 1 ± i√3
=> x = 2 ± i√3.
Now, x³ = 1 when x³ - 1 = (x - 1)(x² + x + 1) = 0, giving x = 1 and:
x² + x + 1 = 0 ==> x = (-1 ± i√3)/2, by the Quadratic Formula.
If we let ω = (-1 - i√3)/2 and ω² = (-1 + i√3)/2, then 1 - 2ω and 1 - 2ω² yield the two complex solutions to (x - 1)³ + 8 = 0. Therefore, the roots of (x - 1)^3 + 8 are -1, 1 - 2ω, and 1 - 2ω²