4B. Determine the set of values of the expression $(x-y)(y-z)(z-x)$ if $x, y$ and $z$ are real numbers for which

$$
\sqrt{x-y+z}=\sqrt{x}-\sqrt{y}+\sqrt{z}
$$
Solution. The given equation can be written in the form

$$
\sqrt{x-y+z}+\sqrt{y}=\sqrt{x}+\sqrt{z}
$$

If we square both sides, we get

$$
\begin{aligned}
& (\sqrt{x-y+z}+\sqrt{y})^{2}=(\sqrt{x}+\sqrt{z})^{2} \\
& 2 \sqrt{y} \sqrt{x-y+z}=2 \sqrt{x} \sqrt{z}
\end{aligned}
$$

If we square again, we get

$$
\begin{aligned}
& y(x-y+z)=x z \\
& x y-y^{2}-(x z-y z)=0 \\
& y(x-y)-z(x-y)=0 \\
& (x-y)(y-z)=0
\end{aligned}
$$

Now it is clear that the value of the expression is equal to

$$
(x-y)(y-z)(z-x)=0(z-x)=0
$$

## III year