This is in continuance to my earlier post : Deriving expression for general cubic equation solution. Also, the links are repeated for the first four pages of the book by Dickson, titled "Introduction to Theory of Algebraic Equations" are given as : page #1,$\,$page #2,$\,$ page #3, $\,$page #4
In the case of general cubic equation of the form $x^3 -c_1x +c_2x^2 -c_4$, am trying to find the significance of the substitutions involved.
a) Page#1 has the first substitution (i) $x = y + \frac{c_1}{3}$,
The original equation has $3$ roots $x_1,x_2,x_3$, with equation as $(x-x_1)(x-x_2)(x-x_3)$ :
$$(x^2 -xx_2-xx_1+x_1x_2)(x-x_3)$$$$= x^3 -x^2(x_1+x_2+x_3)+x(x_1x_3+x_1x_2+x_2x_3)-x_1x_2x_3$$
As, the transformation is linear, i.e. a horizontal shift is involved towards the left by $\frac{c_1}{3}$; so there will be still the same ($3$) number of roots with shift involved on the horizontal axis. Need eliminate the second order term, with coefficient as the sum of roots. Also, by vieta's theorem get from the given cubic $\sum_{i=1}^3 x_i = c_1$. So, the transformed equation (after substitution) has eliminated the needed term: $$x_i = y_i + \frac{c_1}{3}\implies \sum_{i=1}^3 x_i = c_1+\sum_{i=1}^3 y_i\implies c_1 = c_1 + \sum_{i=1}^3 y_i\implies \sum_{i=1}^3 y_i = 0$$
b) Page#2 has the second substitution (i) $y = z - \frac{p}{3z}$, with the two parts making up the root $y$. So, there will be a total of $6$ roots for $3$ values. Also, the substitution can be viewed as : $p= 3z^2 - 3zy\implies p=p_1+p_2$ with $\frac{p_1}{3} =z^2, -\frac{p_2}{3}= zy$. The first is an equation of a circle with radius $\sqrt{\frac{p_1}{3}}$, while the second is that of an hyperbola.
I am unable to make any further logic out of the above.
*Update - * The substitution of (a) is not new, but is also used in quadratic equations by removing the coefficient of $x$ term by substituting in the general quadratic eqn. $ax^2+bx+c=0\implies x^2 +\frac{b}{a}x+\frac{c}{a}=0,$ the value $y = x +\frac{b}{2a}$ to get $y^2 +p =0$, with $p = \frac c a -\frac {b^2}{4a^2}$.
Eqn. of concern for (b) is depressed cubic, with $y$ being the horizontally shifted root. Here, $p$ is the coefficient of the $y$ term, i.e. $p = y_1y_2+y_2y_3+ y_3y_1$. Also, $y_i$ is split into two parts $u_i,v_i$ with product of $3$ valid combinations (out of $1*3+1*3+1*3=3+3+3=9$ possible) of two roots ($u_i, v_j, i,j \in \{1,2,3\}$) leading to $-\frac{p}{3}$. There is logic behind it that concerns restraining by conditions the possible values taken by the roots $u_i, v_j, i,j \in \{1,2,3\}$. The conditions are imposed after grouping the roots $u_i, v_j$ or simply $u,v$, as follows:
$y^3 +py +q =0$ with $p = c_2-\frac{c_1^2}{3}, q= -c_3+\frac13c_1c_2-\frac2{27}c_1^3$
$$(u+v)^3+py +q\implies (u^3+v^3+3u^2v+3uv^2)+p(u+v)+q$$
$\implies u^3+v^3+(3uv+p)(u+v)+q$ with addtl. condn. $3uv + p=0$
Although (on page #41 of the thesis stated below) arbitrary choice is ascribed to the choice of factoring & the addtln. condn. of the last eqn., but logic can be found as follows:
(i) The system of linear equations given by the substitution always yields a solution in $C$, i.e.: $$u+v=y$$$$3uv + p = 0$$
This is as given on footnote on page #41 of the thesis freely available here (that also is the context of the book 'The Unattainable Attempt to Avoid the Casus Irreducibilis for Cubic Equations') by 'Sara Confalonieri', with image here.
As $p = c_2 -\frac{c_1^2}{3}\implies p = \frac{3(x_1x_2+x_2x_3+x_3x_1) -(x_1+x_2+x_3)^2}{3}\implies \frac{-(x_1(x_1+x_2) +x_2(x_2 +x_3)+x_3(x_3+x_1))}{3}.$
Also, $y = x-\frac{c_1}{3}\implies y = x-\frac{x_1+x_2+x_3}{3}.$
Although, still not explored the link between $u,v$ and $p$, but a substitution by an actual value set is done :
For example, the equation set with $y=10, p=-120\implies \frac{-p}{3} = 40$ leads to: $$u + v = 10$$$$uv = 40$$ is equivalent to the equation $u(10-u) =40 \implies u^2 -10u +40 =0$.
(ii) The restriction $3uv + p=0$ leads to further elimination of the first order terms, hence enabling finding the principal cube roots from which the other two cube roots have to be found out.
To continue ...
