The quick answer to that is that this is a well-known example of the second type of irreducible cubics. A more sophisticated answer is that the splitting field can be shown to be Q(a_1, d) where a_1 is any root and d is a square root of the discriminant D = -4p^3 - 27 q^2 In this case D =81 so d= 9 is in Q, so the splitting field is just Q(a_1) and we can see we are in the second case.
Why did you choose x³-3x-1=0 ?
The quick answer to that is that this is a well-known example of the second type of irreducible cubics. A more sophisticated answer is that the splitting field can be shown to be Q(a_1, d) where a_1 is any root and d is a square root of the discriminant
D = -4p^3 - 27 q^2
In this case D =81 so d= 9 is in Q, so the splitting field is just Q(a_1) and we can see we are in the second case.