A simple 2 grade equations system If we have:
$$x^2 + xy + y^2 = 25 $$
$$x^2 + xz + z^2 = 49 $$
$$y^2 + yz + z^2 = 64 $$
How do we calculate $$x + y + z$$
 A: There are solutions $(x,y,z)=(-5,0,8), \ (5,0,-8)$ having different sums $3,-3$, and two others I found on maple, numerically roughly $(-2.2,-3.5,-5.6)$ with sum around $-11.35$, and the same with all positive signs. So my guess is there's not a clever way to get at the sum, however the thing is at worst quadratic for the third answer.
ADDED: By subtracting equations and factoring, one can get three expressions for $x+y+z$ of the type constant over a difference of two of the coordinates. Setting these equal in pairs one finds that either some two of the varibles are equal (I didn't check that leads nowhere), or else in each case the same relation $8x-13y+5z=0$ results. Solving this for $y$ and plugging into the equations, and messing around a bit, leads to an equation for $z$ which factors as $(z-8)(z+8)(129z^2-4096).$ So besides the two integer solutions there may be others for which $z=\pm (64/\sqrt{169})$.
Going back to the other equations then gives $(x,y,z)=(25,40,64)/\sqrt{129}$ [notation meaning deivide each by the radical] as a solution, as well as its opposite obtained by making all signs negative. Looking at other combinations for values of $x,y$ from the equations (given this $z$) did not produce other solutions, in agreement with what maple found.
A: $$x^2 + xy + y^2 = 25 \dots (1)$$
$$x^2 + xz + z^2 = 49 \dots(2)$$
$$y^2 + yz + z^2 = 64 \dots(3)$$
$(2)-(1)$ 
$x(z-y)+(z+y)(z-y)=24$
$\Rightarrow (x+y+z)(z-y)=24$
Similarly we get ,
$(x+y+z)(y-x)=15$ by $(3)-(2)$
$(x+y+z)(z-x)=39$ by $(3)-(1)$
Clearly Let $1/\lambda=(x+y+z)\ne 0$
Then we have, 
$(z-y)=24\lambda$
$(y-x)=15\lambda$
$\Rightarrow x+z-2y=9\lambda$
$\Rightarrow 3y=1/\lambda+9\lambda\Rightarrow y=1/3\lambda+3\lambda$
Now solve for x and put in the first equation to find the value of $\lambda$.
