Hello, chadlyter!
The problems are easier in reverse order.Find all eighth-roots of 1.
Use this to express f (x) = x^8 -1 as a product of irreducible polynomials
a) in C[x] . . b) in R[x] . . c) in Q[x]
(c) Factor the polynomial.
x^8 - 1 .= .(x^4 - 1)(x^4 + 1) .= .(x² - 1)(x² + 1)(x^4 + 1)
. . = .(x - 1)(x + 1)(x² + 1)(x^4 + 1)
(b) The quartic factor can be factored:
. . . . . . . . . . . . . . . . . . . . . . - . . - . ._ . . . . . . . . . _
x^8 - 1 .= .(x - 1)(x + 1)(x² + 1)(x² - √2x + 1)(x² + √2x + 1)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
(a) The eight roots are: . ±1, ±i, (±1 ± i)/√2
To save me typing all those ugly radicals,
. . let: .x1 .= .(1 + i)/√2, .x2 .= .(1 - i)/√2, .x3 .= .(-1 + i)/√2, .x4 .= .(-1 - i)/√2
x^8 - 1 .= .(x - 1)(x + 1)(x - i)(x + i)(x - x1)(x - x2)(x - x3)(x - x4)