$x^2 + y^2 -2xycos(\alpha)-a^2=0$
$y^2 + z^2 -2yzcos(\beta)-b^2=0$
$z^2 + x^2 -2zxcos(\gamma)-c^2=0$
Is there a way to solve this system of equation,yet keep the final expression to some extend short and neat.There could be 8 sets of x y z and only 2 of them are correct.(only real and positive root are considered),How to vadidate them?
The problem might be a little simpler than you think. Note that the first three terms on the left represent the distance between two vectors of length $x$ and $y$ with an angle $\alpha$ between them. $$x^2+y^2-2xy\cos(\alpha)=|\vec{x}-\vec{y}|^2$$ Thus, given the constants, we geometrically conclude that if a solution exists for $x$, $y$ and $z$ then the absolute values of each must form a triangle. Thus, there are $2^3=8$ for a valid set of constants. The constants themselves must satisfy that the sum of the angle is a multiple of $\pi$ and that $a$, $b$ and $c$ also form a triangle.