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The following formulae are a bit similar to cyclotomic polynomials.

\begin{equation*} \prod_{\textsf{all signs}} (x \pm y \pm z) = \left(x^2 + y^2 + z^2\right)^2 \space - \space 4\left(x^2y^2 + x^2z^2 + y^2z^2\right) \end{equation*} \begin{equation*} \prod_{\textsf{all signs}} (w \pm x \pm y \pm z) = \left(\left(w^2 + x^2 + y^2 + z^2\right)^2 \space - \space 4\left(w^2x^2 + w^2y^2 + w^2z^2 + x^2y^2 + x^2z^2 + y^2z^2\right)\right)^2 \space - \space 64w^2x^2y^2z^2 \end{equation*} \begin{equation*} \prod_{0 \le i \le 2} (x + \zeta^i y + \zeta^{2i} z) = x^3 + y^3 + z^3 \space - \space 3xyz \end{equation*} \begin{equation*} \prod_{0 \le i,j \le 2} (x + \zeta^i y + \zeta^j z) = \left(x^3 + y^3 + z^3\right)^3 \space - \space 27x^3y^3z^3 \end{equation*} It is easily established, using CAS, that under the interchange \begin{equation*} \Large{x \enspace \leftrightarrow \enspace e} \end{equation*}

The eliminant for $y^2 = K(x-e_1)(x-e_2)$ is skew-symmetric \begin{equation} \label{eq:e2} \space \frac {\displaystyle \prod_{\textsf{all signs}} \begin{vmatrix} 1 & \hphantom{\pm} \sqrt{(x_1 - e_1)(x_1 - e_2)} \\ 1 & \pm \sqrt{(x_2 - e_1)(x_2 - e_2)} \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 \\ 1 & x_2 \\ \end{vmatrix}} \space = \space (x_1 + x_2) \space - \space (e_1 + e_2) \end{equation}

The eliminant for $y^2 = K(x-e_1)(x-e_2)(x-e_3)$ is symmetric ! \begin{equation} \label{eq:e3} \space \frac {\displaystyle \prod_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & \hphantom{\pm} \sqrt{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)} \\ 1 & x_2 & \pm \sqrt{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)} \\ 1 & x_3 & \pm \sqrt{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)} \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^2} \space = \space \big[(x_1x_2 + x_1x_3 + x_2x_3) - (e_1e_2 + e_1e_3 + e_2e_3)\big]^2 \space - \space 4 (x_1x_2x_3 - e_1e_2e_3)\big[(x_1+x_2+x_3)-(e_1+e_2+e_3)] \end{equation}

The eliminant for $y^2 = K(x-e_1)(x-e_2)(x-e_3)(x-e_4)$ is symmetric !! \begin{equation} \label{eq:e4} \frac {\displaystyle \prod_{\textsf{all signs}} \begin{vmatrix} 1 & x_1 & x_1^2 & \hphantom{\pm} \sqrt{(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)(x_1 - e_4)} \\ 1 & x_2 & x_2^2 & \pm \sqrt{(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)(x_2 - e_4)} \\ 1 & x_3 & x_3^2 & \pm \sqrt{(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)(x_3 - e_4)} \\ 1 & x_4 & x_4^2 & \pm \sqrt{(x_4 - e_1)(x_4 - e_2)(x_4 - e_3)(x_4 - e_4)} \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \\ \end{vmatrix}^4} \enspace = \enspace (x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)^4(e_1+e_2+e_3+e_4)^4 \enspace + \enspace \dots \enspace = \enspace ? \end{equation} It is not hard to see that \eqref{eq:e4} is invariant under Möbius transforms applied simultaneously to the $x_i$ and $e_i$. Therefore it is a simultaneous invariant (in the sense of invariant theory) of degree 8 of the two 4th degree polynomials $\prod_i (t-x_i)$ and $\prod_i (t-e_i)$. It doesn't appear to have a simple formula in terms of the standard 8 generators for $V_4 \oplus V_4$, see [1] pg 88. The Gröbner basis computation needed to compute the formula, runs out of memory, so I can't give it. But because there are 3 generators of degree 2, 4 of degree 3, and 1 of degree 4, it could have up to $15+30+7=52$ terms. So it could also be calculated by inverting a $52 \times 52$ integer matrix, but that's likely to end up with incomputably huge coefficients - ie. not worth attempting. Of course there could be some other simple transvectant formula, which is not a standard generator, tried a few guess's with no success.

The resultant must also be a polynomial in the above invariants, presumably a linear combination of $I_2^2 J_2^2, \space I_2 J_2 K_2^2, \space K_2^4, \space I_3 J_3 K_2, \space I_2 J_3 K_3 + I_3 J_2 K_3', \space K_2 K_3 K_3', \space I_2 J_2 K_4, \space K_2^2 K_4, \space K_4^2$, but my attempts to compute it are not working ...

That suggests that there is a Möbius transform mapping the $x_i$ to the $e_i$. However that is only true if the $j$-invariants are equal, which is easily expressed as a degree 12 transvectant formula \begin{equation*} J_2^3 \thinspace I_3^2 \enspace - \enspace I_2^3 \thinspace J_3^2 \enspace = \enspace 0 \end{equation*} So the cross-ratio's are only equal for a 2-dimensional subspace of points.

The eliminant for $y^3 = K^2(x-e_1)^2(x-e_2)^2(x-e_3)^2$ is skew-symmetric !!! \begin{equation} \label{eq:c3} \frac {\displaystyle \prod_{0 \le i,j \le 2} \begin{vmatrix} 1 & x_1 & \hphantom{\zeta^0} \sqrt[3]{K(x_1 - e_1)(x_1 - e_2)(x_1 - e_3)} \\ 1 & x_2 & \zeta^i \sqrt[3]{K(x_2 - e_1)(x_2 - e_2)(x_2 - e_3)} \\ 1 & x_3 & \zeta^j \sqrt[3]{K(x_3 - e_1)(x_3 - e_2)(x_3 - e_3)} \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^3} \enspace = \enspace (x_1x_2+x_1x_3+x_2x_3)^3(e_1+e_2+e_3)^3 \enspace + \enspace \dots \enspace = \enspace \tfrac 1 {256} \transvectant{\transvectant{X,\transvectant{X,R}_2}_2,\transvectant{R,\transvectant{R,X}_2}_2}_1 \end{equation} where $\zeta$ is a primitive third root of unity and $X(t) = (t-x_1)(t-x_2)(t-x_3)$ and $R(t) = K(t-e_1)(t-e_2)(t-e_3)$.

Equation \eqref{eq:c3} is the simultaneous invariant of two cubic polynomials of degree 6, see [1] pg 83.

Why are they [skew] symmetric?

The eliminant for $y^4 = K(x-e_1)^2(x-e_2)^3(x-e_3)^3$ is \begin{equation} \label{eq:c4} \frac {\displaystyle \prod_{0 \le i,j \le 4} \begin{vmatrix} 1 & \hphantom{\zeta^{0}}\sqrt{(x_1 - e_2)/(x_1 - e_3)} & \hphantom{\zeta^{0}}\sqrt[4]{(x_1 - e_1)^2(x_1 - e_2)/(x_1 - e_3)^3} \\ 1 & \zeta^{2i}\sqrt{(x_2 - e_2)/(x_2 - e_3)} & \zeta^{i}\sqrt[4]{(x_2 - e_1)^2(x_2 - e_2)/(x_2 - e_3)^3} \\ 1 & \zeta^{2j}\sqrt{(x_3 - e_2)/(x_3 - e_3)} & \zeta^{j}\sqrt[4]{(x_3 - e_1)^2(x_3 - e_2)/(x_3 - e_3)^3} \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^4} \space = \space ? \end{equation} where $\zeta$ is a primitive fourth root of unity.

The eliminant for $y^6 = K(x-e_1)^3(x-e_2)^4(x-e_3)^5$ is \begin{equation} \label{eq:c6} \frac {\displaystyle \prod_{0 \le i,j \le 6} \begin{vmatrix} 1 & \hphantom{\zeta^{00}}\sqrt{(x_1 - e_1)/(x_1 - e_3)} & \hphantom{\zeta^00}\sqrt[3]{(x_1 - e_2)/(x_1 - e_3)} \\ 1 & \zeta^{3i}\sqrt{(x_2 - e_1)/(x_2 - e_3)} & \zeta^{2i}\sqrt[3]{(x_2 - e_2)/(x_2 - e_3)} \\ 1 & \zeta^{3j}\sqrt{(x_3 - e_1)/(x_3 - e_3)} & \zeta^{2j}\sqrt[3]{(x_3 - e_2)/(x_3 - e_3)} \\ \end{vmatrix}} {\begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \\ \end{vmatrix}^6} \space = \space ? \end{equation} where $\zeta$ is a primitive sixth root of unity.

References

[1] M. Popoviciu Draisma 2014 Invariants of binary forms. Doctoral Dissertation: Philosophisch-Naturwissenschaftlichen Fakultät der Universität Basel