Factoring $x^3-3x-1$ in terms of $\alpha$ unknown 
I never got a satisfactory answer here: Factoring $x^3-3x-1\in \Bbb Q[x]$ in terms of a unknown root
But all context is contained below

I want to factor $x^3-3x-1\in \Bbb Q[x]$ in terms of an unknown root $\alpha$, i.e. over $\Bbb Q(\alpha)$.
So since $\alpha$ is a root of $x^3-3x-1$ we can divide it out, obtaining:
$(x-\alpha)(x^2+\alpha x+(\alpha^2 -3))$ and obtaining $\alpha^3-3\alpha-1=0$.
Now my next thought would be to apply the quadratic formula to the quadratic factor, but this doesn't yield anything useful, atleast not seemingly, even with me manipulating what I believe is the key, that $\alpha^3-3\alpha-1=0$
We can do other manipulations:
$$x^2+\alpha x+(\alpha^2-3))=x^2+\alpha x + \frac1\alpha=0\implies \alpha x^2 +\alpha^2 x +1 = 0$$
$$\implies \frac{-\alpha^2\pm \sqrt{\alpha^4-4\alpha}}{2\alpha}=-\frac{\alpha}{2}\pm \frac{\sqrt{3\alpha^2-3\alpha}}{2\alpha}$$
Now I am lost again. I am meant to get the other two roots as:
$$\alpha^2-\alpha-2,\quad 2-\alpha^2$$

OK, here's one way to extend your approach to reach the answer needed.
Suppose $a$ is a root.  Then $a^3 = 3a+1$.  This implies we can write higher powers of $a$ as quadratics of $a$:
$$a^4 = 3a^2+a, \quad a^5 = a^2+9a+3, \quad a^6 = 9a^2+6a+1, \quad \dots$$
Further $x^3-3x-1 = (x-a)(x^2+ax+a^2-3) \implies$ the other roots are $r_{\pm} = \dfrac{-a\pm \sqrt{12-3a^2}}2$, as you noted.  
One way from here is to write out a series expansion for the square root term and reduce all the higher order terms to quadratics to simplify.  However often there is an easier way, with a bit of manipulation:
$$12-3a^2 = 16-4(1+3a)+4(3a^2+a)-15a^2+8a \\ =16+8a-15a^2-4(a^3)+4(a^4) \\= (2a^2-a-4)^2$$
So $r_{\pm} = \dfrac{-a\pm (2a^2-a-4)}2 \implies r_+ = a^2-a-2, \quad r_-= 2-a^2$. 
