Solve the following equation

$$
\frac{1}{x+y-z}+\frac{1}{x-y+z}+\frac{1}{-x+y+z}=\frac{1}{x+y+z}
$$

for all its solutions.
Our equation makes sense if

$$
x+y-z \neq 0, \quad x-y+z \neq 0, \quad -x+y+z \neq 0, \quad x+y+z \neq 0
$$

Our equation can also be written as

$$
\begin{aligned}
\left(\frac{1}{z+(x-y)}+\right. & \left.\frac{1}{z-(x-y)}\right)+\left(\frac{1}{(x+y)-z}-\frac{1}{(x+y)+z}\right)= \\
& =\frac{2 z}{z^{2}(x-y)^{2}}+\frac{2 z}{(x+y)^{2}-z^{2}}=0
\end{aligned}
$$

Bringing the two terms on the left side to a common denominator

$$
\begin{aligned}
2 z \cdot & \frac{x^{2}+2 x y+y^{2}-z^{2}+z^{2}-x^{2}+2 x y-y^{2}}{\left[z^{2}-(x-y)^{2}\right]\left[(x+y)^{2}-z^{2}\right]}= \\
& =\frac{8 x y z}{\left[z^{2}-(x-y)^{2}\right]\left[(x+y)^{2}-z^{2}\right]}=0
\end{aligned}
$$

Since according to (1) the denominator is always a finite quantity different from 0, therefore

$$
x y z=0
$$

If $x=0$, according to (1) $y+z \neq 0, y-z \neq 0$, which means $y \neq \pm z$. The same statement holds if $y=0$ or $z=0$. In summary, we can state that all solutions of our equation are: one of the unknowns must necessarily be 0, while the other two are any two numbers - different in absolute value.