What is \( \frac{x^2-y^2-z^2-2 y z}{x^2+y^2-z^2+2 x y}\) + \(\frac{x^2-y^2-z^2-2 y z}{x^2-y^2+z^2-2 x z}\) equal to ?

Given:

 \( \frac{x^2-y^2-z^2-2 y z}{x^2+y^2-z^2+2 x y}\) + \(\frac{x^2-y^2-z^2-2 y z}{x^2-y^2+z^2-2 x z}\) 

Formula Used:

1. (a + b)² = a² + b² + 2ab

2. (a - b)² = a² + b² - 2ab

3. (a² - b²) = (a - b)(a + b)

Calculation:

 \( \frac{x^2-y^2-z^2-2 y z}{x^2+y^2-z^2+2 x y}\) + \(\frac{x^2-y^2-z^2-2 y z}{x^2-y^2+z^2-2 x z}\) 

⇒ \( \frac{x^2-(y^2+z^2+2 y z)}{(x^2+y^2+2 x y)-z^2}\) + \(\frac{x^2-(y^2+z^2+2 y z)}{(x^2+z^2-2 x z)-y^2}\)

By using the identity (1) & (2)

⇒  \( \frac{x^2-(y+z)^2}{(x+y)^2-(z)^2}\) +  \(\frac{x^2-(y+z)^2}{(x-z)^2- (y)^2}\) 

By using the identity 3) 

⇒\(\frac{(x + y + z)(x-y-z)}{(x+y-z)(x+y+z)}\) + \(\frac{(x-y-z)(x+y+z)}{(x-z+y)(x-y-z)}\)

⇒ \(\frac{(x-y-z)}{(x+y-z)}\) + \(\frac{(x+y+z)}{(x-z+y)}\)

⇒ \(\frac{x-y-z+x+y+z}{x+y-z}\)= \(\frac{2x}{x+y-z}\)

∴ The correct answer is \(\frac{2x}{x+y-z}\).