How would I prove this

Let g(x)=x^6-8 ϵ Z[x]

i) Factor g(x) into irreducibles over R
ii) Factor g(x) into irreducible over Q
iii) Show that E= Q(sqrt(-3),sqrt(2)) is the splitting field of g(x) over Q.
iv) Identify up to isomorphism the Galois group of g(x) over Q

i and ii) By difference of cubes,
g(x) = (x^2 - 2)(x^4 + 2x^2 + 4); each of which is easily checked to be irreducible over Q (by rational root theorem for instance)

To factor this over R, use difference of squares:
g(x) = (x^2 - 2)(x^4 + 4x^2 + 4 - 2x^2)
.......= (x^2 - 2)((x^2 + 2)^2 - 2x^2)
.......= (x - √2)(x + √2) * (x^2 + 2 + x√2)(x^2 + 2 - x√2)
(As seen below, the other roots are not real.)

i/iii) Setting each factor equal to 0,
x^2 - 2 = 0 ==> x = ±√2

x^4 + 2x^2 + 4 = 0
==> (x^2 + 1)^2 = -3, by completing the square
==> x^2 = -1 ± i√3 = 2(-2 ± 2i√3)/4
==> x = √2 * (1 ± i√3)/2, -√2 * (1 ± i√3)/2

So, E = Q(√2, i√3) is the splitting field for g(x) over Q.
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iv) Since [E : Q] = 4, Gal(E/Q) is a group of order 4. Via conjugation on each square root 9each of order 2), Gal(E/Q) is isomorphic to the Klein 4-Group.

I hope this helps!