What is the most general form of the cubic matrix equation $X - A = X^{-1}B (X^{-1}BX^{-1}+ C)^{-1}$ that has a real solution of the form $X = f(A,B,C)$, where $A,B$ and $C$ are positive definite matrices? I mean ''general'' to mean: ''what is the smallest amount of structure that you can impose upon $A,B$ and $C$ such that you can give a formula which has $X$ equal to some function of the coefficients.
Obviously, if one assumes that enough of $A$,$B$, and $C$ are diagonal, a solution follows fairly easily.
This problem arises in a model of Bayesian updating, specifically, the right hand side arises from the update of a Gaussian distribution $\textrm{cov}(x,y)\textrm{var}(y)^{-1}$ having some factor structure ($B$ and $C$ are covariance matrices).
Thank you in advance.
Note that $X$ and $B$ are invertible. Then the equation can be rewritten
(1) $X^2-XA+X^2CXB^{-1}X-XACXB^{-1}X-BXB^{-1}X=0,\det(X)\not=0$.
Note that (1) is an equation of degree $4$ and not $3$. Assume $n=2$. When $A,B,C$ are randomly chosen (generic case) in the set of SPD matrices, there are $25$ complex solutions ; among these solutions, there are (in general) between $7$ and $11$ real solutions. Moreover, when solving (1), the key is to find the roots of a polynomial $P$ of degree $25$, the Galois group of which being $S_{25}$ ; thus $P$ is not solvable by radicals and there is no formula $X=f(A,B,C)$.
NB: if a specialization of $A,B,C$ in $M_2(Q)$ leads to the resolution of a polynomial with Galois group $S_{25}$, then in the generic case, the Galois goup $G$ of the obtained polynomial admits $S_{25}$ as a subgroup. In other words, if a specialization leads to a non-solvable polynomial, then the problem is generically non-solvable.