Consider three, rather generic, equations in three variables $x,y,z$: $$ \begin{cases} y z+x=a\\ z x+y=b\\ x y+z=c \end{cases} $$ Now, assuming that there is a solution to this system, is it possible to find an expression in terms of $a,b,c$ only for the sum and/or product of the solutions $x,y,z$?
By this I mean something along the lines of $x+y+z=a+b+c$.
Edit I do not mean the system is true for all values $x,y,z$ as was suggested in a comment below, but that it is true for some set ${x,y,z}$ and the sum and product of these specific values may be expressible in terms of $a,b,c$. Apologies for any confusion...
Some brief experimentation with Mathematica gave me the results $x+y+z=a+b+c$ and $xyz=(a-bc)(b-ca)(c-ab)$ but I'm not sure if this is valid, nor how to make the algebra work to get there.
Secondly... if these solutions are valid, does this make the system symmetric in the group-theory sense? The original system looks to be only partially symmetric, but the solution sub and product actually looks symmetric. I'm not sure if that is significant.
Let $e_1=a+b+c,e_2=ab+ca+bc,e_3=abc$ be the elementary symmetric polynomials in $a,b,c$. Then by resultant computations in Singular I have found that $t=x+y+z$ is the root of a quintic with coefficients in $\mathbb Q[e_1,e_2,e_3]$: $$t^5-e_1t^4+(e_2-4e_1-6)t^3+(4e_1^2-e_3-e_2+14e_1+8)t^2+(-4e_2e_1-8e_1^2+6e_3+3e_2-12e_1-3)t+((e_2+2e_1)^2-3(3e_3+e_2-e_1))=0$$ Adding the three original equations together and subtracting $e_1$ yields $u=xy+zx+yz=e_1-t$. Now the second elementary symmetric function of the original equations' LHSs gives $$xyz=\frac{e_2-u(t+1)}{t-3}$$