Cyclic system of equations Consider the system of equations
$$
\begin{align*}
x^2+(1-y)^2&=a\\
y^2+(1-z)^2&=b\\
z^2+(1-x)^2&=c\\
\end{align*}
$$
Compute $x(1-x)$ in terms of $a,b,c$.
Edit:
The question should say

Compute all possible values of $x(1-x)$ in terms of $a,b,c$

 A: Let $a,b,c$ be $5/2,5/4,1/4.$ Then there are two solutions of the system giving different values for $x(1-x)$, so that one cannot in general determine it from the equations. The two solutions are $(x,y,z)=(3/2,1/2,0)$ and $(x,y,z)=(1/2,-1/2,0).$ Then note that $x(1-x)$ is $-3/4$ for the first solution, but is $+1/4$ for the second solution.
Perhaps the OP has omitted some conditions, maybe that the variables and/or the constants $a,b,c$ are to be integers. However as stated the problem can't have a solution for arbitrary $a,b,c$ to get $x(1-x)$ in terms of them.
Added: Restricting $a,b,c$ to be integers is still not enough to allow computation of $x(1-x)$ from the equations and the values of $a,b,c$. Given integers $m,n$ each of the choices
$$(x,y,z)=(m,\ n+1,\ m+1),\\ (x,y,z)=(-m,\ n+1,\ -m+1)\tag{*}$$
leads to the same values for $a,b,c$ namely
$$(a,b,c)=(m^2+n^2,\ m^2+(n+1)^2,\ 2m^2+2).$$
But the first solution of $(*)$ gives $x(1-x)=m(1-m)=m-m^2,$ while the second solution gives $x(1-x)=-m(1+m)=-m-m^2$ which differs from the first (for any $m \neq 0$).
A: Your question is, "Given the system,
$$x^2+(1-y)^2 = a\tag1$$
$$y^2+(1-z)^2 =b\tag2$$
$$z^2+(1-x)^2 =c\tag3$$
compute all possible values of $x(1-x)$ in terms of $a,b,c$."

Answer: In general, there are eight possible values for $x(1-x)$.

Proof: You have three equations in three unknowns $x,y,z$, hence you can resolve this to one equation in one unknown. Eliminate $y$ between $(1)$ and $(2)$ (easily done with Mathematica's Resultant[] function), then eliminate $z$ between that and $(3)$, and you get the octic in $x$,
$$\small \prod \big(c-\big(b-t+(x-1)^2\color{red}+2\sqrt{t}\pm2\sqrt{b-t-1\color{red}+2\sqrt{t}}\big)\big) \\ \small\big(c-\big(b-t+(x-1)^2\color{red}-2\sqrt{t}\pm2\sqrt{b-t-1\color{red}-2\sqrt{t}}\big)\big)\tag4$$ 
where $t=a-x^2$. Expanding $(4)$, the four sign changes will result in an octic in $x$ with coefficients in $a,b,c$. Since you don't require that it be in radicals, then one can always numerically solve for $x$ for any given $a,b,c$ and, in general, will find eight values for $x(1-x)$.
