How to solve such a system of quadratic equations: $x^2+y^2-xy=a^2, x^2+z^2-xz=b^2, y^2+z^2-yz=c^2$ I don't know how to solve this this system:
$$x^2+y^2-xy=a^2\\x^2+z^2-xz=b^2\\y^2+z^2-yz=c^2$$ The system of quadratic equation in symmetry form has many geometric meaning,
this seems to be a triangular pyramid, three adjacent angles with a common point are $\frac{\pi}{3}$, and opposite to this point is a triangle with $a,b,c$.

But this geometric meaning makes no sense to me, it can't simplify the problem. I don't have other idea about it.

 A: \begin{align}
(y+z-x)(y-z)&=y^2-z^2-xy+xz=a^2-b^2\\
(x+y-z)(x-y)&=x^2-y^2-zx+zy=b^2-c^2\\
(z+x-y)(z-x)&=z^2-x^2-yz+yx=c^2-a^2
\end{align}
Let $u:=y-z$, $v:=y+z$, $\alpha^2:=a^2-b^2$, $\beta^2:=b^2-c^2$, $\gamma^2:=c^2-a^2$, then
\begin{align}
(v-x)u&=\alpha^2\\
(x+u)(u+v-2x)&=-2\beta^2\\
(x-u)(v-u-2x)&=2\gamma^2
\end{align}
Eliminating $x=v-\frac{\alpha^2}{u}$, (unless $\alpha=0=u$) \begin{align}
(uv+u^2-\alpha^2)(u^2+uv-2uv+2\alpha^2)&=-2\beta^2u^2\\
(uv-u^2-\alpha^2)(uv-u^2-2uv+2\alpha^2)&=2\gamma^2u^2\\
\therefore\ (u^2+uv-\alpha^2)(u^2-uv+2\alpha^2)&=-2\beta^2u^2\\
(u^2-uv+\alpha^2)(u^2+uv-2\alpha^2)&=2\gamma^2u^2\\
\therefore\ (u^2+\tfrac{1}{2}\alpha^2)^2-(uv-\tfrac{3}{2}\alpha^2)^2&=-2\beta^2u^2\\
(u^2-\tfrac{1}{2}\alpha^2)^2-(uv-\tfrac{3}{2}\alpha^2)^2&=2\gamma^2u^2
\end{align}
These two equations are actually the same: $$ (u^2-\tfrac{1}{2}\alpha^2)^2-(u^2+\tfrac{1}{2}\alpha^2)^2=-2\alpha^2u^2=2(\beta^2+\gamma^2)u^2$$
So we get $$v=\left(\tfrac{3}{2}\alpha^2\pm\sqrt{(u^2+\tfrac{1}{2}\alpha^2)^2+2\beta^2u^2}\right)/u$$
It seems we've lost one degree of freedom. This is due to taking differences of the equations to start with. So reintroduce the $y,z$ equation as $$(u+v)^2+(u-v)^2+(u+v)(u-v)=4c^2$$ $$3u^2+v^2=4c^2$$ Substituting $v$ in this equation gives a quartic in $u^2$, which can be solved.
Hence find $v$, $y$, $z$, $x$ in turn.
Note: There may be 8 or 4 or 0 real solutions.
A: Solve the first and second equations as quadratic in $y$ and $z$ to get four pairs of solutions
$$y=\frac{1}{2} \left(x-\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
   \left(x-\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x-\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
   \left(x+\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x+\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
   \left(x-\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x+\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
   \left(x+\sqrt{4 b^2-3 x^2}\right)$$
Suppose that the last is the "good" one, replace in the third equation which is
$$(a^2+b^2-c^2)-\frac{5 x^2}{4}+\frac x 4\left(\sqrt{4 a^2-3 x^2}+\sqrt{4 b^2-3 x^2} \right)-\frac{1}{4} \sqrt{4 a^2-3 x^2} \sqrt{4 b^2-3 x^2}=0$$
I suppose that multiple squaring steps would lead to an octic equation in $x$ which, I hope, could be factorized.
However, for given values of $(a,b,c)$, I would prefer to graph the function to locate more or less the solution and after to use Newton method for the solution in $x$ and then $y$ and $z$.
