Solve the system of $3$ quadratic equations Consider the system of equation
$${{x}^{2}}+{{(1-y)}^{2}}=a\\
{{y}^{2}}+{{(1-z)}^{2}}=b\\
{{z}^{2}}+{{(1-x)}^{2}}=c$$
Compute $x(1-x)$ in terms of $a,b,c$.
 A: Contrary to some of the comments, this is not a high degree system.  It turns out to be quadratic.
Geometrically, the system involves three cylinders whose axes are skew perpendicular to one another and equidistant from one another.  With this arrangement there is a threefold symmetry among the cylinder axes, with the symmetry axis lying along line $x=y=z$.  The presence of this symmetry axis leads to multiple simultaneous cancellations in the algebraic treatment.  The cylinders may be unequal in radii so they do not conform to the threefold symmetry, but the parameters that determine the radii do not add any higher degree terms.  Thus the degree of the system is greatly reduced by the symmetry among the axes of the cylinders.
Because the system of equations pairs each variable $x,y,z$ with $1-x,1-y,1-z$, let us consider an origin shift defined by
$(u,v,w)=(x-(1/2),y-(1/2),z-(1/2))$
In terms of the shifted variables we then have the following:
$u^2+v^2+u-v=a-(1/2)\text{.....Eq. 1}$
$v^2+w^2+v-w=b-(1/2)\text{.....Eq. 2}$
$w^2+u^2+w-u=a-(1/2)\text{.....Eq. 3}$
If we add up these equations we discover that the linear terms cancel out producing a sphere centered at the shifted origin.  Here the sum is divided by $2$ to isolate the sum $u^2+v^2+w^2$:
$u^2+v^2+w^2=((a+b+c)/2)-(3/4)\text{.....Eq. 4}$
Now take twice Eq. 1 and subtract Eqs. 2 and 3.  Then similarly take twice Eq. 3 and subtract Eqs. 1 and 2.  Dividing both linear combinations by a common factor of 3 on the left side then gives the following differences between the unknowns:
$u-v=((2a-b-c)/3)\text{.....Eq. 5}$
$w-u=((2c-a-b)/3)\text{.....Eq. 6}$
Now it's easy, albeit messy.  Use Eqs. 5 and 6 to eliminate $v$ and $w$, substitute into Eq. 4 for $u$, and (if I have done it right) get the following:
$108u^2+72(c-a)u+(20(a^2+c^2)-32ac-8b(a-b+c)-18(a+b+c)+27)=0$
This may he solved for $u$ and the original target function, $x(1-x)$, may be rendered as
$((1/2)+u)((1/2)-u)=(1/4)-u^2$.
Note that this will be a unique value only if $a=c$ (the cylinders with axes not parallel to the $x$ axis are congruent) or in the degenerate case if a double root (tangency).
