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:
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:
Updated On: Sep 19, 2024
- \(a + b + c\)
- \(a^2 + b^2 + c^2\)
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The Correct Option is C
Solution and Explanation
The correct option is (C):0
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