Question

If $\frac{x}{(b + c)(b + c - 2a)} = \frac{y}{(c - a)(c + a - 2b)} = \frac{z}{(a - b)(a + b - 2c)}$, then the value of $x + y + z$ is:

• A. $a + b + c$
• B. $a^2 + b^2 + c^2$
• C. 0
• D. 1

Answer 1

Answer: (C)

0

Answer 2

The correct option is (C):0