If ` a - (1/a) = b, b - (1 / b) = c and c - (1 / c) = a ` then what is the value of ` (1 / ab) + (1 / bc) + (1 / ca)` ?

Options

A. `-3`
B. `-6`
C. `-1`
D. `-9`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
\[ a - \frac{1}{a} = b \]
\[ b - \frac{1}{b} = c \]
\[ c - \frac{1}{c} = a \]

2. **Rearranging the Equations**:
We can rearrange each equation to express \(b\), \(c\), and \(a\) in terms of \(a\), \(b\), and \(c\) respectively:
\[ b = a - \frac{1}{a} \]
\[ c = b - \frac{1}{b} = \left(a - \frac{1}{a}\right) - \frac{1}{\left(a - \frac{1}{a}\right)} \]
\[ a = c - \frac{1}{c} \]

3. **Finding the Expression**:
We need to find the value of:
\[ \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} \]

4. **Using the Identity**:
From the rearranged equations, we can derive a useful identity. We know:
\[ a - \frac{1}{a} + b - \frac{1}{b} + c - \frac{1}{c} = 0 \]
This leads to:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = a + b + c \]

5. **Squaring the Identity**:
Now, squaring both sides:
\[ \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2 = (a + b + c)^2 \]
Expanding both sides gives:
\[ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + 2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]

6. **Setting Up the Equation**:
From the earlier steps, we can set up the equation:
\[ 0 = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - 6 + 2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) \]

7. **Solving for the Desired Expression**:
Rearranging gives:
\[ 2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) = 6 - \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right) \]
Since we know that \(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 6\), we can substitute:
\[ 2\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) = 6 - 6 = 0 \]
Thus:
\[ \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = 0 \]

8. **Final Answer**:
Therefore, the value of \(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\) is:
\[ \boxed{-3} \]