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

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{1}{(a-b)(b-c)} - \frac{1}{(b-c)(c-a)} - \frac{1}{(c-a)(a-b)} \]

Given that \( a - b = \frac{1}{b - c} \), we can use this relationship to simplify our calculations.

### Step 1: Substitute \( a - b \)

From the given equation, we have:

\[ a - b = \frac{1}{b - c} \]

Now, we can express \( (a-b)(b-c) \):

\[ (a-b)(b-c) = \left(\frac{1}{b-c}\right)(b-c) = 1 \]

### Step 2: Substitute into the expression

Now substituting \( (a-b)(b-c) = 1 \) into the original expression:

\[ \frac{1}{1} - \frac{1}{(b-c)(c-a)} - \frac{1}{(c-a)(a-b)} \]

This simplifies to:

\[ 1 - \frac{1}{(b-c)(c-a)} - \frac{1}{(c-a)(a-b)} \]

### Step 3: Find common terms

Next, we need to find a common denominator for the last two fractions. The common denominator will be \((b-c)(c-a)(a-b)\).

### Step 4: Rewrite the fractions

Rewriting the fractions with the common denominator:

\[ 1 - \left(\frac{(a-b) + (b-c)}{(b-c)(c-a)(a-b)}\right) \]

### Step 5: Simplify the numerator

The numerator becomes:

\[ 1 - \frac{(a-b) + (b-c)}{(b-c)(c-a)(a-b)} \]

Now, we can simplify \( (a-b) + (b-c) \):

\[ (a-b) + (b-c) = a - c \]

### Step 6: Substitute back

Substituting this back into our expression gives:

\[ 1 - \frac{(a-c)}{(b-c)(c-a)(a-b)} \]

### Step 7: Simplify further

Since we know that \( a - b = \frac{1}{b - c} \), we can also express \( (c-a) \):

\[ c - a = - (a - c) \]

This leads us to:

\[ 1 - \frac{(a-c)}{(b-c)(- (a-c))(a-b)} \]

### Step 8: Final simplification

Notice that the \( (a-c) \) terms cancel out:

\[ 1 + \frac{1}{(b-c)(a-b)} \]

### Conclusion

Thus, the final value of the expression is:

\[ 1 \]