Express one root of depressed cubic equation via another and square root of discriminant Given a cubic polynomial $f(x)=x^3+px+q, p,q\in \mathbb{Q}$ and one of its roots $x_1$, how to express another root $x_2$ in terms of $x_1$, square root of the discriminant $d=\sqrt{-4p^3-27q^2}$, and $p,q$?
I am looking for an expression without radicals. Of course, one can solve $x_1+x_2+x_3=0, x_1 x_2 x_3=-q$ for $x_2,x_3$ and get $x_{2,3}=(-x_1\pm \sqrt{x_1^2+4q/x_1})/2$, but there should exist a relation without square root, because the splitting field of $f(x)$ is $\mathbb{Q}(x_1,x_2)=\mathbb{Q}(x_1,d)$, and hence $x_2$ is a linear combination $x_2=ax_1+bd$ with some $a,b\in \mathbb{Q}(x_1,x_2)$.
 A: Here is a set of formulas that works :
$$
\begin{array}{lcl}
x_2 &=& -\frac{x_1}{2}-\frac{2d}{4p^3+27q^2}
\bigg(p^2-\frac{9q}{4}x_1+\frac{3p}{2}x_1^2\bigg) \\
x_3 &=& -\frac{x_1}{2}+\frac{2d}{4p^3+27q^2}
\bigg(p^2-\frac{9q}{4}x_1+\frac{3p}{2}x_1^2\bigg) \\
\end{array}\tag{1}
$$
It is easy to see that the solutions must be of the form
$-\frac{x_1}{2}+g(x_1)d$ where $g$ is some polynomial, but I see no 
natural way to derive the formula for $g(x_1)$.
To check that (1) is  correct, it suffices to notice the identity
$$
\bigg(p^2-\frac{9q}{4}x+\frac{3p}{2}x^2\bigg)^2=
(4p^3+27q^2)\frac{3x^2+4p}{16}+\frac{9p}{4}(x^3+px+q)(px-3q)
\tag{2}
$$
A: Here is a solution I found. In the formulas for $x_{2,3}$ the term $\sqrt{x_1^2+4q/x_1}=\delta$ is a square root of the discriminant of the quadratic equation for $x_{2,3}$:
 $$g(z)=(z-x_2)(z-x_3)=z^2+x_1 z-q/x_1=0.$$ On the other hand, by definition of discriminant $\delta=(x_2-x_3)$ and $d=(x_1-x_2)(x_1-x_3)(x_2-x_3)$. Therefore  $\delta=d/(x_1-x_2)(x_1-x_3)=d/g(x_1)$, which gives us
$$x_{2,3}=(-x_1\pm d/g(x_1))/2=(-x_1\pm d/(3x_1^2-q/x_1))/2.$$
