If ` a + (1)/( b) = b + (1)/( c) = c = (1)/( a) ( a ne b ne c)` then the value of abc is

Options

A. `pm 1`
B. `pm 2 `
C. 0
D. `pm (1)/(2)`

Answer

To solve the equation \( a + \frac{1}{b} = b + \frac{1}{c} = c = \frac{1}{a} \) where \( a \neq b \neq c \), we can follow these steps:

### Step 1: Set the common value
Let \( k = a + \frac{1}{b} = b + \frac{1}{c} = c = \frac{1}{a} \).

### Step 2: Express \( c \) in terms of \( k \)
From the equation \( c = k \).

### Step 3: Express \( a \) in terms of \( k \)
From \( \frac{1}{a} = k \), we can express \( a \) as:
\[ a = \frac{1}{k} \]

### Step 4: Substitute \( c \) into the second equation
Substituting \( c = k \) into \( b + \frac{1}{c} = k \):
\[ b + \frac{1}{k} = k \]
This leads to:
\[ b = k - \frac{1}{k} \]

### Step 5: Write down all variables in terms of \( k \)
Now we have:
- \( a = \frac{1}{k} \)
- \( b = k - \frac{1}{k} \)
- \( c = k \)

### Step 6: Calculate \( abc \)
Now, we need to find \( abc \):
\[ abc = \left(\frac{1}{k}\right) \left(k - \frac{1}{k}\right) (k) \]
This simplifies to:
\[ abc = \frac{1}{k} \cdot k \cdot \left(k - \frac{1}{k}\right) = (k - \frac{1}{k}) \]

### Step 7: Simplify the expression
Now, we can simplify \( abc \):
\[ abc = k - \frac{1}{k} \]

### Step 8: Find the value of \( k \)
From the earlier equations, we also know that:
\[ k - 1 = \frac{1}{k} \]
Multiplying both sides by \( k \):
\[ k^2 - k - 1 = 0 \]
Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ k = \frac{1 \pm \sqrt{5}}{2} \]

### Step 9: Calculate \( abc \)
Now substituting \( k \) back into \( abc \):
Using \( k = \frac{1 + \sqrt{5}}{2} \):
\[ abc = \left(\frac{1 + \sqrt{5}}{2}\right) - \frac{2}{1 + \sqrt{5}} \]

### Step 10: Rationalizing the denominator
Rationalizing \( \frac{2}{1 + \sqrt{5}} \):
\[ \frac{2(1 - \sqrt{5})}{(1 + \sqrt{5})(1 - \sqrt{5})} = \frac{2(1 - \sqrt{5})}{-4} = \frac{1 - \sqrt{5}}{-2} = \frac{\sqrt{5} - 1}{2} \]
Thus:
\[ abc = \frac{1 + \sqrt{5}}{2} - \frac{\sqrt{5} - 1}{2} = 1 \]

### Final Answer:
Thus, the value of \( abc \) is \( 1 \).

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