If ` x= b + c - 2a , y = c + a - 2 b , z = a + b -2 c ` then the value of ` x^(2) + y^(2) - z^(2) + 2xy ` is

Options

A. 0
B. a+ b + c
C. a - b + c
D. a + b - c

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 - z^2 + 2xy \) given the expressions for \( x \), \( y \), and \( z \):

1. **Given Expressions**:
\[ x = b + c - 2a \]
\[ y = c + a - 2b \]
\[ z = a + b - 2c \]

2. **Calculate \( x + y \)**:
\[ x + y = (b + c - 2a) + (c + a - 2b) \]
Simplifying this:
\[ = b + c - 2a + c + a - 2b \]
\[ = (b - 2b) + (c + c) + (-2a + a) \]
\[ = -b + 2c - a \]

3. **Calculate \( x + y + z \)**:
\[ x + y + z = (b + c - 2a) + (c + a - 2b) + (a + b - 2c) \]
Simplifying this:
\[ = b + c - 2a + c + a - 2b + a + b - 2c \]
\[ = (b - 2b + b) + (c + c - 2c) + (-2a + a + a) \]
\[ = 0 \]

4. **Substituting into the Expression**:
Now, we substitute \( x + y + z = 0 \) into the expression \( x^2 + y^2 - z^2 + 2xy \):
\[ x^2 + y^2 + 2xy - z^2 = (x + y)^2 - z^2 \]
Using the difference of squares:
\[ = (x + y + z)(x + y - z) \]
Since \( x + y + z = 0 \):
\[ = 0 \cdot (x + y - z) = 0 \]

5. **Final Result**:
Therefore, the value of \( x^2 + y^2 - z^2 + 2xy \) is:
\[ \boxed{0} \]