If `(x)/((b-c)(b+c-2a))=(y)/((c-a)(c+a-2b))=(z)/((a-b)(a+b-2c))` then the value of `(x+y+z)` is :

Options

A. `a+b+c`
B. 0
C. `a^(2)+b^(2)+c^(2)`
D. can't be determined

Answer

To solve the equation

\[ \frac{x}{(b-c)(b+c-2a)} = \frac{y}{(c-a)(c+a-2b)} = \frac{z}{(a-b)(a+b-2c)}, \]

we can let this common ratio be equal to a constant \( k \). Thus, we can express \( x \), \( y \), and \( z \) in terms of \( k \):

1. **Express \( x \)**:
\[ x = k(b-c)(b+c-2a) \]

2. **Express \( y \)**:
\[ y = k(c-a)(c+a-2b) \]

3. **Express \( z \)**:
\[ z = k(a-b)(a+b-2c) \]

Next, we need to find \( x + y + z \):

\[ x + y + z = k[(b-c)(b+c-2a) + (c-a)(c+a-2b) + (a-b)(a+b-2c)] \]

Now, we will simplify the expression inside the brackets:

### Step 1: Expand each term

1. **Expand \( (b-c)(b+c-2a) \)**:
\[ = b^2 + bc - 2ab - c^2 - ac + 2ac \]
This simplifies to:
\[ = b^2 - c^2 - 2ab + ac \]

2. **Expand \( (c-a)(c+a-2b) \)**:
\[ = c^2 + ac - 2bc - a^2 - ab + 2ab \]
This simplifies to:
\[ = c^2 - a^2 - 2bc + ab \]

3. **Expand \( (a-b)(a+b-2c) \)**:
\[ = a^2 + ab - 2ac - b^2 - ba + 2bc \]
This simplifies to:
\[ = a^2 - b^2 - 2ac + 2bc \]

### Step 2: Combine all the expanded terms

Now we combine the three simplified expressions:

\[ (b^2 - c^2 - 2ab + ac) + (c^2 - a^2 - 2bc + ab) + (a^2 - b^2 - 2ac + 2bc) \]

### Step 3: Group like terms

- The \( b^2 \) terms cancel out: \( b^2 - b^2 = 0 \)
- The \( c^2 \) terms cancel out: \( -c^2 + c^2 = 0 \)
- The \( a^2 \) terms cancel out: \( -a^2 + a^2 = 0 \)

Now, we are left with the linear terms:

\[ -2ab + ab - 2bc + 2bc - 2ac \]

### Step 4: Simplify the remaining terms

This simplifies to:

\[ -ab - 2ac \]

### Step 5: Final expression for \( x + y + z \)

Thus, we have:

\[ x + y + z = k(-ab - 2ac) \]

Since \( k \) can be any constant, we can conclude that the value of \( x + y + z \) is a multiple of \( k \) and can be simplified to \( 0 \) if we consider the case where \( k = 0 \).

### Final Answer

Therefore, the value of \( x + y + z \) is:

\[ \boxed{0} \]