I'm trying to understand the technique used by Lagrange to solve cubic and quartic equations. I have read that the Lagrange resolvent for the cubic is
$$ x_1+\omega x_2+ \omega^2 x_3 $$
where $\omega$ is the principal cubic root of 1.
My question is: Why isn't the resolvent for the quartic
$$ x_1+\omega x_2 +\omega^2 x_3 +\omega^3 x_4 $$
where $\omega$ is the principal quartic root of 1?
Why did Lagrange use $x_1-x_2+x_3-x_4$?
Is there an intuitive explanation?
More generally, is there an intuitive way to understand Lagrange resolvent?
It is. The lagrange resolvent of a quartic is by definition $x_1 + \omega x_2 + \omega^2 x_3 + \omega^3 x_4$. But this is just a naming convention to honor Lagranges achievements.
One can solve the quartic using $x_1 + \omega x_2 + \omega^2 x_3 + \omega^3 x_4$. And Lagrange did this. See my answer here to understand how. Doing this, one necessary crosses a resolvent which is only fixed by the klein four group $V_4$ and has three images under the $4!=24$ permutations of the roots.
Lagrange noticed that there are more, easier such resolvents, which have this probability and which can be used to solve the quartic, like
$$x_1 x_2 + x_3 x_4,$$ $$(x_1 + x_2)(x_3 + x_4),$$ or $$(x_1 + x_2-x_3-x_4)^2.$$