TITLE: Logic behind cubic resolution of $x^4+px^2+qx+r=0$
QUESTION [7 upvotes]: I'm reading "A course in algebra" by E. B. Vinberg to have a basic understanding of algebra. 
In its Chapter 3 there's an interesting demonstration on cubic resolution of equation $x^4+px^2+qx+r=0$.

Define $\sigma_i$ the elementary symmetric polynomials:
$$\sigma_i=\sum_{k_1<k_2<\cdots<k_i}x_{k_1}x_{k_2}\cdots x_{k_i}$$
Define $h_i(x_1,x_2,x_3,x_4)$:
$$ \left\{
     \begin{array}{ll} 
     h_1=x_1x_2+x_3x_4 \\ 
     h_2=x_1x_3+x_2x_4\\
     h_3=x_1x_4+x_2x_3
     \end{array}
     \right.$$
Then one has:
$$ \left\{
     \begin{array}{ll}
       h_1+h_2+h_3&=\sigma_2 \\
       h_1h_2+h_1h_3+h_2h_3&=\sigma_1\sigma_3-4\sigma_4\\
       h_1h_2h_3&=\sigma_1^2\sigma_4+\sigma_3^2-4\sigma_2\sigma_4
     \end{array}
   \right. $$
Let $c_1$, $c_2$, $c_3$, $c_4$ be the roots of equation $x^4+px^2+qx+r=0$.
Then 
$$ \left\{
     \begin{array}{ll} 
     \sigma_1(c_1,c_2,c_3,c_4)=0\\ 
     \sigma_2(c_1,c_2,c_3,c_4)=p\\
     \sigma_3(c_1,c_2,c_3,c_4)=-q\\
     \sigma_4(c_1,c_2,c_3,c_4)=r
     \end{array}
     \right.$$
Let $d_i=h_i(c_1,c_2,c_3,c_4)$.
So $d_i$ are the roots of equation $y^3+a_1y^2+a_2y+a_3=0$ where 
$$ \left\{
     \begin{array}{ll} 
     a_1=-(d_1+d_2+d_3)\\ 
     a_2=d_1d_2+d_1d_3+d_2d_3\\
     a_3=-d_1d_2d_3
     \end{array}
     \right.$$    
From step 2 one has $a_1=-p$, $a_2=-4r$, $a_3=4pr-q^2$.
So the equations has the form $y^3-py^2-4ry+(4pr-q^2)=0$. This is assumed solvable.
Define
$$ \left\{
     \begin{array}{ll} 
     b_1=c_1+c_2-c_3-c_4\\ 
     b_2=c_1-c_2+c_3-c_4\\ 
     b_3=c_1-c_2-c_3+c_4
     \end{array}
     \right.$$    
So one has
$$ \left\{
     \begin{array}{ll} 
     b_1^2=4(d_1-p)\\ 
     b_2^2=4(d_2-p)\\ 
     b_3^2=4(d_3-p)\\ 
     b_1b_2b_3=-8q
     \end{array}
     \right.$$    
From here it leads to
$$ \left\{
     \begin{array}{ll} 
     c_1+c_2+c_3+c_4&=0\\ 
     c_1+c_2-c_3-c_4&=b_1&=2\sqrt{d_1-p_1}\\ 
     c_1-c_2+c_3-c_4&=b_2&=2\sqrt{d_2-p_1}\\ 
     c_1-c_2-c_3+c_4&=b_3&=2\sqrt{d_3-p_1}
     \end{array}
     \right.$$    
So it's solved:
$$c_{1,2,3,4}=\frac{1}{2}(\pm\sqrt{d_1-p_1}\pm\sqrt{d_2-p_1}\pm\sqrt{d_3-p_1})$$
, with number of minuses being even, and the value of the square roots chosen so that $b_1b_2b_3=-8q$.

Now, all these steps I understand, and they are quite smart. However, if we look at step 5, is it just some math trick (being sensitive to math forms) or there's something, some higher logic behind  it? 
The logic before step 5, to put it simple, is to get $d_i$ via $\sigma_i$, using the symmetry, without knowing exactly $c_i$. This is quite smart. So step 5 the idea is to get $c_i$ from $d_i$.
How ppl get the formula in step 5 to achieve that? is it just by try and error, by luck?
Or, there's some logic behind it? I suspect so as $b_i$ look similar to $d_i$, there seems also some "indirect symmetry" behind the construction of $b_i$.
Please throw some lights?

REPLY [4 votes]: There wasn't any luck. The hardest part was actually in coming up with $d_1,d_2,d_3$. After that it is just a matter of manipulation to make use of the fact that $c_1 + c_2 + c_3 + c_4 = 0$ because the quartic is depressed.
$d_1 = c_1 c_2 + c_3 c_4 = p - ( c_1 c_3 + c_1 c_4 + c_2 c_3 + c_2 c_4 ) = p - (c_1+c_2)(c_3+c_4) = p + (c_1+c_2)^2$
$d_2 = c_1 c_3 + c_2 c_4 = p - ( c_1 c_2 + c_1 c_4 + c_3 c_2 + c_3 c_4 ) = p - (c_1+c_3)(c_2+c_4) = p + (c_1+c_3)^2$
$d_3 = c_2 c_3 + c_1 c_4 = p - ( c_1 c_2 + c_1 c_3 + c_4 c_2 + c_4 c_3 ) = p - (c_1+c_4)(c_2+c_3) = p + (c_1+c_4)^2$
From here it is clear that we can get $c_1,c_2,c_3,c_4$ because:
$c_1+c_2 = \sqrt{d_1-p}$ where the sign of the $\sqrt{}$ is unknown
$c_1+c_3 = \sqrt{d_2-p}$ where the sign of the $\sqrt{}$ is unknown
$c_1+c_4 = \sqrt{d_3-p}$ where the sign of the $\sqrt{}$ is such that:
  $\sqrt{d_1-p} \sqrt{d_2-p} \sqrt{d_3-p} = c_1^2 (c_1+c_2+c_3+c_4) + (c_1c_2c_3+c_1c_2c_4+c_1c_3c_4+c_2c_3c_4)$
  $ = -q$ because $c_1 + c_2 + c_3 + c_4 = 0$
$c_1 + c_2 + c_3 + c_4 = 0$ (again)
The sum of the first three minus the fourth gives:
$2 c_1 = \sqrt{d_1-p} + \sqrt{d_2-p} + \sqrt{d_3-p}$
You can check that your textbook arranged the above steps in quite a backward way, which doesn't help in understanding where the various expressions came from.
Notes
Similar manipulations also give rise to the fact that the resolvent cubic $y \mapsto y^3 − p y^2 - 4r y + (4pr−q^2)$ has the same discriminant as the original quartic.
$d_1 - d_2 = ( c_1 c_2 + c_3 c_4 ) - ( c_1 c_3 + c_2 c_4 ) = (c_1-c_4)(c_2-c_3)$
$d_1 - d_3 = ( c_1 c_2 + c_3 c_4 ) - ( c_1 c_4 + c_2 c_3 ) = (c_1-c_3)(c_2-c_4)$
$d_2 - d_3 = ( c_1 c_3 + c_2 c_4 ) - ( c_1 c_4 + c_2 c_3 ) = (c_1-c_2)(c_3-c_4)$
Then it is natural to consider the product of both sides which gives the result.
Alternatives
There is another resolvent cubic where the above steps are actually much easier to see:
Let $d_1 = - (c_1+c_2)(c_3+c_4) = (c_1+c_2)^2$
Let $d_2 = - (c_1+c_3)(c_2+c_4) = (c_1+c_3)^2$
Let $d_3 = - (c_1+c_4)(c_2+c_3) = (c_1+c_4)^2$
Then by symmetry the symmetric polynomials on $(d_1,d_2,d_3)$ will be symmetric in $(c_1,c_2,c_3,c_4)$ as well, and from those it is easy to compute the resolvent cubic that has $(d_1,d_2,d_3)$ as roots. After that the solution is identical, and in some sense simpler than the one your textbook uses. Later on it will also turn out that the criterion to distinguish whether the Galois group of an irreducible separable quartic is $C_4$ or $D_4$ will be much simpler using this resolvent cubic than your textbook's.