`x,y,z` are distinct real numbers such that `x+1/y = y + 1/z =z + 1/x` The value of `x^(2)y^(2)z^(2) `is………..

Answer

To solve the problem where \( x, y, z \) are distinct real numbers such that

\[ x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x}, \]

we want to find the value of \( x^2 y^2 z^2 \).

### Step 1: Set the common value
Let \( k = x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} \).

This gives us three equations:
1. \( x + \frac{1}{y} = k \)
2. \( y + \frac{1}{z} = k \)
3. \( z + \frac{1}{x} = k \)

### Step 2: Rearranging the equations
From these equations, we can express \( x, y, z \) in terms of \( k \):
1. \( x = k - \frac{1}{y} \)
2. \( y = k - \frac{1}{z} \)
3. \( z = k - \frac{1}{x} \)

### Step 3: Substituting to find relationships
Substituting \( y \) from the second equation into the first:
\[ x = k - \frac{1}{k - \frac{1}{z}}. \]
This can be simplified further, but let's also consider the other equations.

### Step 4: Subtracting equations
Subtract the first equation from the second:
\[ (y - x) = \left(\frac{1}{z} - \frac{1}{y}\right). \]
This simplifies to:
\[ y - x = \frac{y - z}{zy}. \]
Rearranging gives us:
\[ (y - x)zy = y - z. \]

### Step 5: Similar equations from other differences
Similarly, we can derive:
1. \( (z - y)zx = z - x \)
2. \( (x - z)xy = x - y \)

### Step 6: Multiply the equations
Now, we multiply the three derived equations:
\[ (x - y)(y - z)(z - x) = \frac{(y - z)(z - x)(x - y)}{xyz}. \]
This leads to:
\[ (x - y)(y - z)(z - x)xyz = (y - z)(z - x)(x - y). \]
Since \( x, y, z \) are distinct, we can cancel out the common terms, leading to:
\[ xyz = 1. \]

### Step 7: Finding \( x^2 y^2 z^2 \)
Now, squaring both sides gives us:
\[ (xyz)^2 = 1^2 \implies x^2 y^2 z^2 = 1. \]

### Final Answer
Thus, the value of \( x^2 y^2 z^2 \) is:

\[ \boxed{1}. \]