Suppose we want to solve the "reduced" quartic equation $x^4+px^2+qx+r=0$ by means of Lagrange resolvent. I denote the roots by $x_1, x_2, x_3, x_4$; we have $x_1+x_2+x_3+x_4=0$.
In many texts one typically reads: the "generic" Lagrange resolvent would be $R=x_1+ix_2-x_3-ix_4$; however, Lagrange found easier resolvents, for example $(x_1+x_2)(x_3+x_4)$.
That's fine; but I still want to see how $R$ works.
When one permutes the roots in all possible ways, $R^4$ has six (rather than three) distinct values. So we get an equation $\prod (x-\ldots)$ of degree 6 (its coefficients are some polynomials in $p,q,r$). At this point some texts say "with some tricks, it can be reduced to degree 3". I computed this equation: its coefficients are not pleasant, and have no clear pattern. So I guess that the tricks should not be applied on the explicit form.
Hence the question: What are the tricks? How one reduces this equation to degree 3?
I tried to query the original work of Lagrange for answers. Even though I do not speak french, I believe I found the solution.
The lagrange resolvent for the quartic is $R = x_1 + i x_2 - x_3 -i x_4$. You can write this as a complex number
$$R = (x_1 - x_3) + i(x_2 - x_4)$$
This lagrange resolvent has $4!=24$ permutations. One of these permutations is
$$ \bar{R} = (x_1 - x_3)-i(x_2 - x_4) $$
the complex conjugate of $R$.
Note that
$$R\bar{R} = (x_1-x_3)^2+(x_2-x_4)^2$$
only has three images under all $24$ permutations of $x_1,x_2,x_3,x_4$:
$$\begin{alignat*}{5} R\bar{R} &={}&(x_1-x_3)^2+(x_2-x_4)^2 &=:{}&t_1 \\ &&(x_1-x_2)^2+(x_3-x_4)^2 &=:{}&t_2 \\ &&(x_1-x_4)^2+(x_2-x_3)^2 &=:{}&t_3 \\ \end{alignat*}$$
These are fixed by the klein four group $V_4$ and can be found as the solution of a cubic polynomial
$$(X-t_1)(X-t_2)(X-t_3).$$