The numerical value of ` (( a - b)^(2))/( ( b - c) ( c - a)) + (( b - c)^(2))/( ( c - a) ( a - b))+ (( c - a) ^(2))/( ( a - b) ( b - c))` is `( a ne b ne c)`

Options

A. 0
B. 1
C. `(1)/(3)`
D. 3

Answer

To solve the expression

\[ \frac{(a-b)^2}{(b-c)(c-a)} + \frac{(b-c)^2}{(c-a)(a-b)} + \frac{(c-a)^2}{(a-b)(b-c)}, \]

we will follow these steps:

### Step 1: Identify the common denominator
The common denominator for the three fractions is

\[ (b-c)(c-a)(a-b). \]

### Step 2: Rewrite each term with the common denominator
We can rewrite each term over the common denominator:

\[ \frac{(a-b)^2 (a-b)}{(b-c)(c-a)(a-b)} + \frac{(b-c)^2 (b-c)}{(c-a)(a-b)(b-c)} + \frac{(c-a)^2 (c-a)}{(a-b)(b-c)(c-a)}. \]

This simplifies to:

\[ \frac{(a-b)^3 + (b-c)^3 + (c-a)^3}{(b-c)(c-a)(a-b)}. \]

### Step 3: Apply the identity for the sum of cubes
Using the identity for the sum of cubes, we know that:

\[ x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz). \]

Here, let \( x = a-b \), \( y = b-c \), and \( z = c-a \). Then we have:

\[ x + y + z = (a-b) + (b-c) + (c-a) = 0. \]

### Step 4: Simplify using the identity
Since \( x + y + z = 0 \), we can use the identity:

\[ (a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a). \]

### Step 5: Substitute back into the expression
Now substituting this back into our expression gives:

\[ \frac{3(a-b)(b-c)(c-a)}{(b-c)(c-a)(a-b)}. \]

### Step 6: Cancel out the common terms
The \((b-c)(c-a)(a-b)\) terms in the numerator and denominator cancel out, resulting in:

\[ 3. \]

### Final Answer
Thus, the numerical value of the given expression is:

\[ \boxed{3}. \]