If `(x)/(b + c - a) = (y)/(c + a - b) = (z)/(a + b - c)` , then value of x(b -c) + y(c -a) + z(a - b) is equal to

Options

A. 1
B. 0
C. 1 + b + c
D. `a^(2) + b^(2) + c^(2)`

Answer

To solve the problem, we start with the given equation:

\[ \frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c} \]

Let us denote this common ratio as \( k \). Therefore, we can express \( x \), \( y \), and \( z \) in terms of \( k \):

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

Now, we need to find the value of the expression:

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

Substituting the values of \( x \), \( y \), and \( z \):

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

Factoring out \( k \):

\[ = k \left[ (b + c - a)(b - c) + (c + a - b)(c - a) + (a + b - c)(a - b) \right] \]

Now, we will expand each term inside the brackets:

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

2. **Second term**:
\[ (c + a - b)(c - a) = c^2 - ac + a^2 - ab - bc + ba = c^2 - a^2 - ab + ac \]

3. **Third term**:
\[ (a + b - c)(a - b) = a^2 - ab + b^2 - bc - ac + cb = a^2 - b^2 - ac + bc \]

Now, combining all three terms:

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

Now, let's simplify:

- The \( b^2 \) and \( -b^2 \) cancel out.
- The \( c^2 \) and \( -c^2 \) cancel out.
- The \( a^2 \) and \( -a^2 \) cancel out.
- The \( -ab \) terms combine to give \( -2ab \).
- The \( ac \) terms combine to give \( 2ac \).
- The \( bc \) term remains.

Putting it all together, we have:

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

Now, we can see that the expression simplifies to zero:

\[ = 0 \]

Thus, the final value of \( x(b - c) + y(c - a) + z(a - b) \) is:

\[ \boxed{0} \]