I'm trying to understand the history of solving polynomial equations and have come across Lagrange's work attempting to generalize the process of finding formulas for quadratics, cubics, and quartics. Here is what I understand so far:
For a quadratic $f(x)$ with roots $x_1, x_2$, he studies the quantity $p(x_1, x_2) = x_1 - x_2$. One observes that it takes two values on permutation of the roots, since $p(x_1, x_2) = -p(x_2, x_1)$. Then one can construct an additional polynomial
$$g(z) = (z-p(x_1, x_2))(z-p(x_2, x_1)) = z^2 - p(x_1, x_2)^2$$
which is symmetric in $x_1, x_2$ (since by construction we include every permutation, and thus permitting $x_1, x_2$ leads to just a shuffling of linear terms). Since $g(z)$ is symmetric in $x_1, x_2$, its coefficients are symmetric in $x_1, x_2$ and thus may be expressed in terms of the elementary symmetric polynomials in $x_1, x_2$ (i.e. the coefficients of our original $f(x)$) by the Fundamental Theorem of Symmetric Polynomials. Furthermore, solving $g(z)=0$–an easier difference of squares then the general quadratic–gives us values for $p(x_1, x_2)$ and $p(x_2, x_1)$ in terms of these elementary symmetric polynomials (i.e. original coefficients of $f(x)$). Then we can solve this linear system to find $x_1, x_2$ in terms of our original coefficients of $f(x)$.
For a cubic $f(x)$, he follows much the same procedure. We now investigate the quantity $p(x_1, x_2, x_3) = x_1 + \omega x_2 + \omega^2 x_3$ where $\omega$ is the cube root of unity, observing it takes six values given the six permutations of $x_1, x_2, x_3$. Then we construct
$$g(z) = \prod (z-p(x_i, x_j, x_k)).$$
a 6th-degree polynomial symmetric in $x_1, x_2, x_3$, but one that happens to be quadratic in $z^3$ which we know how to solve. In fact, this is why a square root shows up in Cardano's method. Again, $g(z)$ has yielded an easier equation to solve, which we call the Lagrange resolvent, that is vital to finding solutions to $f(x)$.
But why has Lagrange chosen to consider specifically the quantities $p(x_1, x_2) = x_1 - x_2$ and $p(x_1, x_2, x_3) = x_1 + \omega x_2 + \omega^2 x_3$? More generally, what is the motivation behind the value:
$$p(x_1, \dots, x_n) = x_1 + \zeta x_2 + \zeta^2 x_3 + \cdots + \zeta^{n-1} x_n$$
where $\zeta$ is the $n$th root of unity? Further, why in the quartic case, does he break this, and instead of $p(x_1, x_2, x_3, x_4) = x_1 + ix_2 - x_3 - ix_4$ why does he consider $p(x_1, x_2, x_3, x_4) = x_1x_2 + x_3x_4$? Section 4 of this document seems to present a Galois-theoretic motivation, but why did Lagrange look towards these expressions?
