Analytical solution for a system of three bivariate polynomials of degree two in three unknowns

Does this system of three nonlinear equations in three unknowns admit a closed-form solution?
If so, which strategy brings to it?

$ x^2 + a_1\, xy + y^2 = a_0$
$ y^2 + b_1\, yz + z^2 = b_0$
$ z^2 + c_1\, zx + x^2 = c_0$

I know that
$|a_1| < 2, |b_1| < 2, |c_1| < 2$
$a_1 \, b_1 \,c_1 \neq 0$
$a_0 > 0, b_0 > 0, c_0 >0$

This is an elaboration on a prior post.

A similar system has been discussed here, which I find hard to link to this more generic question.

Not a mathematician.

Variables can be eliminated between polynomial equations using resultants. For the system of equations here, eliminating $\,y,z\,$ between the three equations results in a polynomial equation of degree $\,8\,$ in $\,x\,$.

However, the particular form of these equations means that if $\,(x,y,z)\,$ is a solution then so is $(-x,-y,-z)$, so the octic in $\,x\,$ turns out to be a bi-quartic i.e. a quartic in $\,x^2\,$. The explicit equation is too unwieldy to derive by hand, but can be easily computed with a CAS, for example this is WA's result.

The general quartic can be solved by radicals, so it is technically possible to work out the closed-form solutions for $\,x\,$, and then $\,y,z\,$ after substituting back into the original equations. This is hardly practical, though, since the quartic here is irreducible in general, so the solutions will carry the full weight of the general quartic formulas.

Particular cases can still lead to simplified calculations, for example $a_1 = -1,$ $b_1 = 0$, $c_1 = 1$, $a_0 = 3$, $b_0 = 2$, $c_0 = 1$ gives $x^8 - 7 x^6 + 18 x^4 - 16 x^2 + 4 = 9$ with some "nice" real roots $\,x=\pm 1, \pm \sqrt{2 - \sqrt[3]{4}}\,$.