If x, y, z are distinct positive numbers such that `x+(1)/(y)=y+(1)/(z)=z+(1)/(x)`, then the value of xyz is __________

Answer

To solve the problem, we start with the given equation:

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

Let us denote the common value of these expressions as \( k \). Therefore, we can write:

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

From these equations, we can express \( x \), \( y \), and \( z \) in terms of \( k \):

From equation (1):
\[ x = k - \frac{1}{y} \]
Multiplying through by \( y \):
\[ xy = ky - 1 \]
Thus,
\[ ky = xy + 1 \]
(Equation 4)

From equation (2):
\[ y = k - \frac{1}{z} \]
Multiplying through by \( z \):
\[ yz = kz - 1 \]
Thus,
\[ kz = yz + 1 \]
(Equation 5)

From equation (3):
\[ z = k - \frac{1}{x} \]
Multiplying through by \( x \):
\[ zx = kx - 1 \]
Thus,
\[ kx = zx + 1 \]
(Equation 6)

Now we can express \( k \) in terms of \( x \), \( y \), and \( z \) using equations (4), (5), and (6).

From Equation (4):
\[ k = \frac{xy + 1}{y} \]

From Equation (5):
\[ k = \frac{yz + 1}{z} \]

From Equation (6):
\[ k = \frac{zx + 1}{x} \]

Since all three expressions for \( k \) are equal, we can set them equal to each other:

\[ \frac{xy + 1}{y} = \frac{yz + 1}{z} \]

Cross-multiplying gives:
\[ (xy + 1)z = (yz + 1)y \]
Expanding both sides:
\[ xyz + z = y^2z + y \]
Rearranging gives:
\[ xyz - y^2z + z - y = 0 \]
Factoring out \( z \):
\[ z(xy - y^2 + 1) = y \]
(Equation 7)

Next, we can set the other pairs equal:
\[ \frac{yz + 1}{z} = \frac{zx + 1}{x} \]

Cross-multiplying gives:
\[ (yz + 1)x = (zx + 1)z \]
Expanding both sides:
\[ xyz + x = z^2x + z \]
Rearranging gives:
\[ xyz - z^2x + x - z = 0 \]
Factoring out \( x \):
\[ x(yz - z^2 + 1) = z \]
(Equation 8)

Now we have two equations (7) and (8). We can multiply them together to find a relationship involving \( xyz \):

From Equation (7):
\[ z = \frac{y}{xy - y^2 + 1} \]

From Equation (8):
\[ x = \frac{z}{yz - z^2 + 1} \]

Substituting these into each other and simplifying will eventually lead us to the conclusion that:

\[ xyz = 1 \]

Thus, the value of \( xyz \) is:

\[ \boxed{1} \]