Field of decomposition of $x^3-x^2-4x-1.$ over $\mathbb{Q}$ Suppose we have $f(x)=x^3-x^2-4x-1$ and $\alpha \in \mathbb{C}$ a root of $f(x)$. (We do not know the value of $\alpha$.) It is easy to see that $f(x)$ is irreducible in $\mathbb{Q}[x]$.
It is also easy to prove $-\frac{1}{1+\alpha}$ is also a root of $f(x)$. I need to show $\mathbb{Q}(\alpha)$ is the decomposition field of $f(x)$. Since $f(x)$ is separable, we have three distinct roots $\alpha,-\frac{1}{1+\alpha}, \gamma$. So the decomposition field is $\mathbb{Q}(\alpha,-\frac{1}{1+\alpha}, \gamma)=\mathbb{Q}(\alpha, \gamma)$.
How do I show that this is actually equal to $\mathbb{Q}(\alpha)$ without actually calculating the roots of $f(x)?$
 A: Notice that the sum of all the roots equals the coefficient at $x^2$, e.g. $-1$. So $\gamma$ can be expressed in polynomial terms of $\alpha$. 
A: You already have two excellent answers by Jef and Will, but I really do think it is worth mentioning a method that you can just memorise and apply mindlessly in future situations.
Let $\alpha$, $\beta$ and $\gamma$ be the three roots of your irreducible cubic. The Galois group is a subgroup of the permutation group on $ \{ \alpha, \beta, \gamma \}$. Furthermore, the Galois group acts transitively on the roots, since the cubic is irreducible. So there are two possibilities:


*

*The Galois group is $S_3$, in which case the degree of the extension is six, so the splitting field is larger than $\mathbb Q(\alpha)$.

*The Galois group is $A_3$, in which case the degree of the extension is three, so the splitting field is equal to $\mathbb Q(\alpha)$.


To distinguish between the two cases, consider the discriminant of the cubic:
$$ \Delta = (\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2.$$


*

*If the Galois group is $S_3$, then the odd permutations in $S_3$ send $\sqrt{\Delta} \mapsto - \sqrt{\Delta}$, so $\sqrt{\Delta}$ is not fixed by the Galois group, hence $\sqrt{\Delta} \notin \mathbb Q$.

*If the Galois group is $A_3$, then the Galois group only contains even permutations, which all send $\sqrt{\Delta} \mapsto + \sqrt{\Delta}$, so $\sqrt{\Delta}$ is fixed by the Galois group, hence $\sqrt{\Delta} \in \mathbb Q$.


There is a formula for the discriminant in terms of the coefficients of the cubic. If the cubic is
$$ f(x) = x^3 + bx^2 + cx + d,$$
then the discriminant is
$$ \Delta = b^2 c^2 - 4c^3 -4b^3 d - 27d^2 + 18bcd. $$
[Note that this simplifies to $\Delta = -4c^3 - 27 d^2$ if $b = 0$, which is easy to remember.]
In your example, $\Delta = 169$, so $\sqrt{\Delta} = 13 \in \mathbb Q$, hence the Galois group is $A_3$, hence the splitting field is $\mathbb Q(\alpha)$.
Needless to say, your method and the answers by Will and Jef show far more ingenuity than what I have suggested, but still I think it's good to have a reliable tool that you can always fall back upon.
A: Let f:=x^3-x^2-4x-1. It is easy to see that f is an irreducible polynomial over Q. In fact, the degree of f is 3 and so if we prove that f hasn't rational roots, we can conclude that f is irreducible (over Q). Since 1 and -1 are not zeros of f, f is then irreducible.
Now, a is a root of f. You can easy proved that b=:-1/(1+a) is also a root of f. And so, the third root is: c= -1/(1+b) or equivalently: -1/((-1)^3 * ab) = -(1+a)/a. Since the degree of f is odd, f has at least one real root. Since f is separable (f is irreducible and the characteristic of Q is zero), f has not multiple roots. And so: a, b and c are three (simple) real roots. Q(a) is then the splitting field of f over Q (because b and c are clearly elements in Q(a)) and Q(a)/Q is a Galois extension. 
I hope this helps you. 
I am very sorry for my bad English. 
