## Task A-1.3.

Let $x, y$ and $z$ be distinct real numbers, none of which are equal to zero, such that

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

Determine the value of the expression $x^{2} y^{2} z^{2}$.
## Solution.

From $x+\frac{1}{y}=y+\frac{1}{z}$, by multiplying with $y z$ we have $x y z+z=y^{2} z+y$, that is,

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

Since $x$ and $y$ are different, dividing by $x-y$ gives

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

From the remaining two equalities, similarly we have

$$
z x=\frac{z-x}{y-z} \quad \text{and} \quad x y=\frac{x-y}{z-x}.
$$

Multiplying now yields

$$
y z \cdot z x \cdot x y=\frac{y-z}{x-y} \cdot \frac{z-x}{y-z} \cdot \frac{x-y}{z-x}
$$

from which $x^{2} y^{2} z^{2}=1$.