I'd like to solve the following system of three quadratic that has obvious symmetric in its structure,
$ x^2 + y^2 + xy = a^2 \tag{1}$
$ y^2 + z^2 + y z = b^2 \tag{2}$
$ z^2 + x^2 + zx = c^2 \tag{3}$
My attempt:
Multiplying $(1)$ by $b^2$ and subtracting $a^2$ times $(2)$ gives the homogeneous equation
$ b^2 x^2 + (b^2 - a^2) y^2 - a^2 z^2 + b^2 xy - a^2 y z = 0 \tag{4} $
Similarly, multiplying $(1)$ by $c^2$ and subtracting $a^2$ time $(3)$ gives
$ (c^2 - a^2) x^2 + c^2 y^2 + c^2 xy - a^2 z^2 - a^2 xz = 0 \tag{5}$
Assuming $x \ne 0$, then dividing $(4)$ and $(5)$ by $x^2$ yields
$ b^2 + (b^2 - a^2) u^2 - a^2 v^2 + b^2 u - a^2 u v = 0 \tag{6} $
$ (c^2 - a^2) + c^2 u^2 + c^2 u - a^2 v^2 - a^2 v = 0 \tag{7} $
where $u = \dfrac{y}{x}$ and $v = \dfrac{z}{x} $
Equation $(7)$ is nice in that it doesn't have the $uv$ term, so completing the squares in $u$ and $v$ gives
$ (c^2 - a^2) + c^2 ( u + \dfrac{1}{2} )^2 - a^2 ( v + \dfrac{1}{2} )^2 = \dfrac{1}{4} (c^2 - a^2) $
which reduces to
$ c^2 ( u + \dfrac{1}{2} )^2 - a^2 ( v + \dfrac{1}{2} )^2 = \dfrac{3}{4} (a^2 - c^2) \tag{8}$
Assuming $a^2 - c^2 \gt 0 $, then the solution of this equation is
$ u = -\dfrac{1}{2} + \dfrac{K}{c} \sec \theta \tag{9} $
$ v = - \dfrac{1}{2} + \dfrac{K}{a} \tan \theta \tag{10}$
where $K^2 = \dfrac{3}{4} (a^2 - c^2) $
Substituting $(9)$ and $(10)$ into $(6)$, gives a trigonometric equation involving $\sec \theta , \tan \theta , \sec^2 \theta, \tan^2 \theta, \sec \theta \tan \theta $. This equation can be transformed into an equation involving just $\cos \theta, \sin \theta, \sin^2 \theta, \cos^2 \theta $ by multiplying it through by $\cos^2 \theta $.
The final shape of equation $(6)$ takes the form
$ A \cos \theta + B \sin \theta + C \cos(2 \theta) + D \sin(2 \theta) + E = 0 $
And there is a standard technique to solve this equation that uses the substitution $ t = \tan\left( \dfrac{\theta}{2} \right)$. This transforms this trigonometric equation into a quartic (degree 4) polynomial in $t$. Solutions $t$ can be found numerically using the well-known Newton's method, or a similar method, or even better using Bairstow's method, or can be computed exactly (also numerically) using the quartic root formulas. If you're using Mathematica or Sage or a similar math app there are functions to compute the roots of any polynomial.
Once the roots $t$ have been computed, the corresponding $\theta$'s can be found by inverting the defining formula, and finally $u, v$ can be computed. And finally $x$ can be computed from the original equations, and so can $y$ and $z$.
That's what I wanted to present as a possible way to solve this particular problem.
Your comments, hints, or additional information, and alternative solutions are appreciated.
Thank you all.