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

Options

A. ` a + b + c `
B. 0
C. ` a^(2) + b^(2) + c^(2)`
D. 2

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 set all three fractions equal to a common variable \( k \). Thus, we have:

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

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

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

Factoring out \( k \):

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

Now, let's simplify the expression inside the brackets:

1. **Expanding each term**:
- For \( (b - c)(b + c - 2a) \):
\[ = b^2 + bc - 2ab - c^2 + ac - 2ac = b^2 - c^2 - 2ab + ac. \]
- For \( (c - a)(c + a - 2b) \):
\[ = c^2 + ac - 2bc - a^2 + ab - 2ab = c^2 - a^2 - 2bc + ab. \]
- For \( (a - b)(a + b - 2c) \):
\[ = a^2 + ab - 2ac - b^2 + ac - 2bc = a^2 - b^2 - 2ac + ab. \]

2. **Combining all terms**:
Now we combine all three expanded terms:
\[ (b^2 - c^2 - 2ab + ac) + (c^2 - a^2 - 2bc + ab) + (a^2 - b^2 - 2ac + ab). \]

When we combine like terms:
- The \( b^2 \) terms cancel with \( -b^2 \),
- The \( c^2 \) terms cancel with \( -c^2 \),
- The \( a^2 \) terms cancel with \( -a^2 \),
- The \( -2ab + ab + ab \) leads to \( 0 \),
- The \( -2bc + ab + ab \) leads to \( 0 \),
- The \( -2ac + ac + ac \) leads to \( 0 \).

Thus, all terms cancel out, resulting in:

\[ (b^2 - b^2) + (c^2 - c^2) + (a^2 - a^2) + 0 + 0 + 0 = 0. \]

Therefore, we have:

\[ x + y + z = k \cdot 0 = 0. \]

So, the final answer is:

\[ \boxed{0}. \]