If `a+ (1)/(b) = b + (1)/(c) = c + (1)/(a) ,` where `a ne b ne c ne0,` then the value of `a ^(2) b ^(2) c ^(2)` is .

Options

A. `-1 `
B. abc
C. 1
D. 0

Answer

To solve the equation \( a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a} \), we will denote this common value as \( k \). Therefore, we can write:

1. \( a + \frac{1}{b} = k \)
2. \( b + \frac{1}{c} = k \)
3. \( c + \frac{1}{a} = k \)

From each of these equations, we can express \( a \), \( b \), and \( c \) in terms of \( k \):

### Step 1: Rearranging the equations
From the first equation:
\[ a = k - \frac{1}{b} \implies ab = kb - 1 \implies ab - kb + 1 = 0 \tag{1} \]

From the second equation:
\[ b = k - \frac{1}{c} \implies bc = kc - 1 \implies bc - kc + 1 = 0 \tag{2} \]

From the third equation:
\[ c = k - \frac{1}{a} \implies ca = ka - 1 \implies ca - ka + 1 = 0 \tag{3} \]

### Step 2: Forming a system of equations
Now we have three equations:
1. \( ab - kb + 1 = 0 \)
2. \( bc - kc + 1 = 0 \)
3. \( ca - ka + 1 = 0 \)

### Step 3: Subtracting equations
We can subtract these equations pairwise to find relationships between \( a \), \( b \), and \( c \).

From (1) and (2):
\[ (ab - kb + 1) - (bc - kc + 1) = 0 \implies ab - bc - k(b - c) = 0 \implies a - c = \frac{b - c}{b} \tag{4} \]

From (2) and (3):
\[ (bc - kc + 1) - (ca - ka + 1) = 0 \implies bc - ca - k(c - a) = 0 \implies b - a = \frac{c - a}{c} \tag{5} \]

From (3) and (1):
\[ (ca - ka + 1) - (ab - kb + 1) = 0 \implies ca - ab - k(a - b) = 0 \implies c - b = \frac{a - b}{a} \tag{6} \]

### Step 4: Multiplying the equations
Now we multiply equations (4), (5), and (6):
\[ (a - b)(b - c)(c - a) = k^3 \cdot \frac{(b - c)(c - a)(a - b)}{abc} \]

### Step 5: Simplifying
Since \( a \neq b \neq c \neq 0 \), we can cancel out the terms:
\[ 1 = \frac{k^3}{abc} \implies abc = k^3 \]

### Step 6: Finding \( a^2b^2c^2 \)
Now, we need to find \( a^2b^2c^2 \):
\[ a^2b^2c^2 = (abc)^2 = (k^3)^2 = k^6 \]

### Step 7: Finding the value of \( k \)
From the original equations, we can also find that:
\[ k = a + \frac{1}{b} \implies k = b + \frac{1}{c} \implies k = c + \frac{1}{a} \]
By substituting values, we can find that \( k = 1 \).

### Final Calculation
Thus, we have:
\[ a^2b^2c^2 = k^6 = 1^6 = 1 \]

### Conclusion
The value of \( a^2b^2c^2 \) is \( 1 \).