I am trying to understand why $$\det(F^{\mu\nu})=(\vec{E}\cdot\vec{B})^2\tag{1}.$$ Of course one can just calculate the determinant of $F^{\mu\nu}$ expressed as a matrix with components given in terms of $E_x, E_y, ...$, etc., but I am looking for something a bit more insightful. In particular, I understand that $\vec{E}\cdot\vec{B}$ can be written as
$$\vec{E}\cdot\vec{B}=-\frac{1}{8}\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}.\tag{2}$$
This seems promising because then
$$(\vec{E}\cdot\vec{B})^2=\frac{1}{64}\epsilon_{\mu\nu\rho\sigma}\epsilon_{\alpha\beta\gamma\delta}F^{\mu\nu}F^{\rho\sigma}F^{\alpha\beta}F^{\gamma\delta},\tag{3}$$
which looks awfully like the expression for the determinant of $F$ in terms of Levi-Civita tensors,
$$\det(F^{\mu\nu})=\frac{1}{4!}\epsilon_{\mu\nu\rho\sigma}\epsilon_{\alpha\beta\gamma\delta}F^{\mu\alpha}F^{\nu\beta}F^{\rho\gamma}F^{\sigma\delta}.\tag{4}$$
But I can't figure out how to connect these two expressions. Since $F$ is antisymmetric, one can flip the two indices of any single $F$ at the cost of a minus sign, but I need a way to permute indices between different $F$'s. I'm also not sure how a factor of 3 could possibly enter in to change the 1/64 into a 1/24.