`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, we start with the given condition:

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

Let's denote this common value as \( k \). Therefore, we can write:

1. \( x + \frac{1}{y} = k \) (Equation 1)
2. \( y + \frac{1}{z} = k \) (Equation 2)
3. \( z + \frac{1}{x} = k \) (Equation 3)

From each equation, we can express \( x, y, z \) in terms of \( k \):

From Equation 1:
\[ x = k - \frac{1}{y} \]
\[ \Rightarrow yx = ky - 1 \]
\[ \Rightarrow ky - yx = 1 \]
\[ \Rightarrow y(k - x) = 1 \]
\[ \Rightarrow y = \frac{1}{k - x} \] (Equation 4)

From Equation 2:
\[ y = k - \frac{1}{z} \]
\[ \Rightarrow zy = kz - 1 \]
\[ \Rightarrow kz - zy = 1 \]
\[ \Rightarrow z(k - y) = 1 \]
\[ \Rightarrow z = \frac{1}{k - y} \] (Equation 5)

From Equation 3:
\[ z = k - \frac{1}{x} \]
\[ \Rightarrow xz = kx - 1 \]
\[ \Rightarrow kx - xz = 1 \]
\[ \Rightarrow x(k - z) = 1 \]
\[ \Rightarrow x = \frac{1}{k - z} \] (Equation 6)

Now, we can set up the equations for \( x, y, z \) using Equations 4, 5, and 6.

Next, we will subtract pairs of equations to derive relationships between \( x, y, z \):

Subtracting Equation 1 from Equation 2:
\[ (y + \frac{1}{z}) - (x + \frac{1}{y}) = 0 \]
\[ \Rightarrow y - x = \frac{1}{y} - \frac{1}{z} \]
\[ \Rightarrow y - x = \frac{z - y}{yz} \]
\[ \Rightarrow (y - x)yz = z - y \]
\[ \Rightarrow y^2z - xyz + y = z \]
This gives us our first equation.

Subtracting Equation 2 from Equation 3:
\[ (z + \frac{1}{x}) - (y + \frac{1}{z}) = 0 \]
\[ \Rightarrow z - y = \frac{1}{x} - \frac{1}{z} \]
\[ \Rightarrow z - y = \frac{z - x}{xz} \]
\[ \Rightarrow (z - y)xz = z - x \]
This gives us our second equation.

Subtracting Equation 3 from Equation 1:
\[ (x + \frac{1}{y}) - (z + \frac{1}{x}) = 0 \]
\[ \Rightarrow x - z = \frac{1}{y} - \frac{1}{x} \]
\[ \Rightarrow x - z = \frac{x - y}{xy} \]
This gives us our third equation.

Now we can multiply these three derived equations together:

1. \( (y - x)(z - y)(x - z) \)
2. \( (z - y)(y - z)(x - y) \)

The left-hand side becomes:
\[ (y - x)(z - y)(x - z) \]

The right-hand side becomes:
\[ \frac{(y - z)(z - x)(x - y)}{(xyz)^2} \]

Now, we can simplify this to find \( x^2y^2z^2 \).

After simplifying, we find that:
\[ x^2y^2z^2 = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]