If `a + (1)/(b) = b + (1)/(c) = c + (1)/(a) ` (where `a ne b ne c),` then , abc is equal to

Options

A. `+1 `
B. `-1`
C. `+1 & -1`
D. None of these

Answer

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

### Step 1: Set up the equations
We can write the following equations based on the given information:
1. \( a + \frac{1}{b} = k \)
2. \( b + \frac{1}{c} = k \)
3. \( c + \frac{1}{a} = k \)

### Step 2: Rearranging the equations
From the first equation, we can express \( a \):
\[ a = k - \frac{1}{b} \quad \text{(1)} \]
From the second equation, we can express \( b \):
\[ b = k - \frac{1}{c} \quad \text{(2)} \]
From the third equation, we can express \( c \):
\[ c = k - \frac{1}{a} \quad \text{(3)} \]

### Step 3: Substitute and simplify
Now, we will substitute equation (2) into equation (1):
\[ a = k - \frac{1}{k - \frac{1}{c}} \]
To simplify \( \frac{1}{k - \frac{1}{c}} \):
\[ \frac{1}{k - \frac{1}{c}} = \frac{c}{kc - 1} \]
Thus, we can rewrite \( a \):
\[ a = k - \frac{c}{kc - 1} \]

Next, substitute equation (3) into equation (2):
\[ b = k - \frac{1}{k - \frac{1}{a}} \]
Simplifying \( \frac{1}{k - \frac{1}{a}} \):
\[ \frac{1}{k - \frac{1}{a}} = \frac{a}{ka - 1} \]
Thus, we can rewrite \( b \):
\[ b = k - \frac{a}{ka - 1} \]

### Step 4: Equating the differences
Now, we can equate the differences:
From \( a + \frac{1}{b} = b + \frac{1}{c} \):
\[ a - b = \frac{1}{c} - \frac{1}{b} \]
This can be rearranged to:
\[ a - b = \frac{b - c}{bc} \]
This gives us our first equation:
\[ (a - b)bc = b - c \quad \text{(4)} \]

Similarly, we can derive:
From \( b + \frac{1}{c} = c + \frac{1}{a} \):
\[ b - c = \frac{1}{a} - \frac{1}{c} \]
This gives us our second equation:
\[ (b - c)ac = c - a \quad \text{(5)} \]

And from \( c + \frac{1}{a} = a + \frac{1}{b} \):
\[ c - a = \frac{1}{b} - \frac{1}{a} \]
This gives us our third equation:
\[ (c - a)ab = a - b \quad \text{(6)} \]

### Step 5: Multiply the equations
Now, we multiply equations (4), (5), and (6):
\[ (a - b)(b - c)(c - a) = \frac{(b - c)(c - a)(a - b)}{abc} \]
This simplifies to:
\[ abc = 1 \]

### Conclusion
Thus, we find that:
\[ abc = \pm 1 \]