nothing. the degree of an extension matches the degree of a polynomial only if the polynomial is IRREDUCIBLE.
let's look at x6 - 1:
this isn't irreducible, it factors as:
x6 - 1 = (x3 - 1)(x3 + 1) = (x - 1)(x + 1)(x2 + x + 1)(x2 - x +1).
so one thing is clear from this factorization, if we adjoin the roots of x2 - x + 1 to Q, we get at least 4 of the roots of x6 - 1.
the 6-th roots of 1 form a cyclic subgroup of the multiplicative group of Q(u)*, where u is a root of x2 - x + 1 (the other root is u5 = 1/u.
it turns out that u + 1/u = 1. you can use trigonometry to prove this: u = 1/2 + (√3/2)i)
we obviously can't have a subgroup of a group of order 6 of order 4: if we have at least 4 elements of it, we have them all.
also, if u is a 6-th root of 1, clearly u2 is a 3-rd root of 1, so u2 is a root of x3 - 1, so we can write:
x6 - 1 = (x - 1)(x + 1)(x - u2)(x - 1/u2)(x - u)(x - 1/u)
(it turns out that u2 + 1/u2 = -1. this can be proved using some complex number algebra, since u2 = -1/2 + (√3/2)i).
another way to look at it:
suppose u is a non-rational root of x3 + 1. this means that u3 = -1.
but -1 is a SQUARE root of 1 (remember every positive number has TWO square roots: the positive one, and the negative one. we often forget about the negative one).
so if u3 = -1, then:
u6 = (u3)2 = (-1)2 = 1, that is: u is a 6-th root of unity.
6 = 3*2, so it is NOT surprising that including the "3 part" (the 6-th roots that come from the cubic x3 + 1), we get everything, because Q already contains the "2-part"
(that is: the factor x2 - 1 of x6 - 1 splits completely over Q).
what IS sort of surprising is this, we have:
Q < Q(u2) ≤ Q(u)
but since [Q(u):Q] = 2, we must have either Q(u2) = Q, or Q(u2) = Q(u).
it turns out that Q(u2) = Q(u). the easiest way to prove this is to show we can write u in terms of u2:
so convince yourself that u = -1/u2, which will prove Q(u) ≤ Q(u2).