8. Solve the system of equations $\left\{\begin{array}{l}x^{2}-y^{2}+z=\frac{27}{x y} \\ y^{2}-z^{2}+x=\frac{27}{y z} \\ z^{2}-x^{2}+y=\frac{27}{x z}\end{array}\right.$.

Answer. $(3 ; 3 ; 3),(-3 ;-3 ; 3),(-3 ; 3 ;-3),(3 ;-3 ;-3)$.
By adding all three equations of the system, we get $x+y+z=\frac{27}{x y}+\frac{27}{y z}+\frac{27}{x z} \Leftrightarrow x+y+z=\frac{27(x+y+z)}{x y z}$. There are two possible cases: $x+y+z=0 \quad$ or $x y z=27$.

a) $x+y+z=0$, then $z=-x-y$. Substituting into the first two equations of the original system:

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } - y ^ { 2 } - x - y = \frac { 2 7 } { x y } } \\
{ y ^ { 2 } - x ^ { 2 } - 2 x y - y ^ { 2 } + x = - \frac { 2 7 } { y ( x + y ) } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+y)(x-y-1)=\frac{27}{x y} \\
x(x+2 y-1)=\frac{27}{y(x+y)}
\end{array}\right.\right.
$$

Dividing the first equation by the second (which is possible because the right side of the second equation is not zero), we get $\frac{(x+y)(x-y-1)}{x(x+2 y-1)}=\frac{x+y}{x}$, from which $x+y=0$ (and then $z=0$) or $x-y-1=x+2 y-1$ (and then $y=0$). Both cases are impossible, as by the domain of definition, the variables are not equal to zero.

b) $x y z=27$. Then $z=\frac{27}{x y}, x=\frac{27}{y z}$, so the first two equations of the system become $x^{2}-y^{2}=0$, $y^{2}-z^{2}=0$, i.e., $|x|=|y|=|z|$. Considering $x y z=27$, we get 4 solutions: $(3 ; 3 ; 3),(-3 ;-3 ; 3),(-3 ; 3 ;-3)$, $(3 ;-3 ;-3)$.