how to find the solution to this system of equations Given the system of equations:
$
xy+xz=54+x^2  \\ 
yx+yz=64+y^2 \\ 
zx+zy=70+z^2 $
Need to find all of the solutions of $ x,y$ and $z$.
Tried to sum up all three equations but got stuck with nothing to factorize. 
 A: First assume $x,y,z\ne 0$.
From equation (1), $y+z=\dfrac {54}{x}+x\Rightarrow y+z-x=\dfrac {54}{x}$     ..........(A1)
From equation (2), $z+x=\dfrac {64}{y}+y\Rightarrow z+x-y=\dfrac {64}{y}$     ..........(A2)
From equation (3), $x+y=\dfrac {70}{z}+z\Rightarrow x+y-z=\dfrac {70}{z}$............(A3)
Now, adding (A1) & (A2), $2xyz=54y+64x$......(B1)
Adding (A2) & (A3), $2xyz=64z+70y$........(B2) 
Adding (A3) & (A1), $2xyz=70x+54z$...........(B3)
Equating (B1) & (B2), $64x-16y-64z=0$
Equating (B2) & (B3), $70x-70y-16z=0$
Equating (B3) & (B1), $16x-54y-54z=0$.
Now solve these equations.
A: with the method above we get $(x,y,z)=\pm (9, 8, 7)$
A: Adding, $x^{2}$ ,$y^{2}$, & $z^{2}$ to 1st,2nd & 3rd equations respectively we get, 
$x(x+y+z)=2x^{2}+54$
$y(x+y+z)=2y^{2}+64$
$z(x+y+z)=2z^{2}+70$
So, $\dfrac{2x^{2}+54}{x}=\dfrac{2y^{2}+64}{y}=\dfrac{2z^{2}+70}{z}=x+y+z=k(say)$
From these, $2x^{2}-kx+54=0,2y^{2}-ky+64=0,2z^{2}-kz+70=0$.
So, $x=\dfrac{k\pm \sqrt{k^{2}-4.2.54}}{4}, y=\dfrac{k\pm \sqrt{k^{2}-4.2.64}}{4}, z=\dfrac{k\pm \sqrt{k^{2}-4.2.70}}{4}$.
Putting, $k=24$,& taking only positive sign we get, $x=9,y=8,z=7$.
Note that, $x+y+z=24$. So, if we take negative sign then $x=3,y=4,z=5$, in that case $x+y+z\ne 24$. So, it is not admissible.
Again, if we take, $k=-24$ then the solutions are $x=-9,y=-8,z=-7$ 7 in this case sum is also $-24$.
A: HINT:

$ x ( x+y+z - 2 x ) =54 \, ;  y  (x-y+z- 2 y )=64 \,;  z (x+y+z- 2 z)=70 $

