Is there a quick way to compute the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$? Is there a way to find the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$? 
Just analyzing the roots shows that the polynomial is separable, so I suppose the splitting field would be a Galois extension. However, the roots are not easy to get a handle on, so it's not obvious to me what roots would need to be adjoined to $\mathbb{Q}$ to get the degree.
What is the right way to do this? Thanks.

Just some culture, in case of curiosity. If you take any prime $q$ that has an integral expression as 
$$ q = 2 x^2 + x y + 4 y^2,  $$ such as $2, \; 5, \; 7\; 19,$ then
$$ z^3 + z + 1 \equiv 0 \pmod q $$ has no roots in integral $z.$
For the special case of $31,$ we have
$$  z^3 + z + 1 \equiv (z -3)(z - 14)^2 \pmod {31}.   $$
For any other prime $$ p =  x^2 + x y + 8 y^2,  $$ such as $47, \; 67, \; 131, \; 149,$ then
$$ z^3 + z + 1 \equiv 0 \pmod p $$ has three distinct roots in integral $z$ and factors as three distinct linear factors.
For any prime $r$ with $(-31|r) = -1,$ such as $3, \; 11, \; 13, \; 17,$ then 
$$ z^3 + z + 1 \equiv 0 \pmod r $$ has a single non-repeated root, so the cubic factors as a linear times a quadratic. 
Well, why not. It turns out that every integer $n,$ positive or negative or $0,$ has an expression in integers as
$$ n =  x^2 + x y + 8 y^2 + z^3 + z,$$ where we deliberately strip off the 1.
The difficult question is, what integers $n$ have an expression in integers as
$$ n = 2 x^2 + x y + 4 y^2 + z^3 + z \; ?$$
Certainly not all, $n = \pm 1$ do not work. The first few, in absolute value, that do not work are $$ \pm 1, \; \pm 869, \; \pm 25171, \; \pm 21118439, \; \pm 611705641,  $$ these being the odd integers $u$ with
$$ 27 u^2 - 31 v^2 = -4.   $$ The first few even values of $u$ are
$$  30 = 3^3 + 3, $$
$$  729090 = 90^3 + 90, $$
$$  17718345150 = 2607^3 + 2607, $$
so these are  easily expressed as  $2 x^2 + x y + 4 y^2 + z^3 + z $ with both $x,y = 0.$
Well, somebody did mention the discriminant of a cubic, we have
$$ \mbox{disc}_z \left( z^3 + z + 1 \right) = -31    $$
and
$$ \mbox{disc}_z \left( z^3 + z + u \right) = -4 - 27 u^2 = -31 v^2.$$
So there.
