Discussion
\(\dfrac{(x^{2}+y^{2}-z^{2})^{2}-(x^{2}-y^{2}+z^{2})^{2}}{(y+z)(z-y)}\) is equal to :
Answer: C
The expression equals \(-4x^{2}\).
The numerator is a difference of two squares, \(a^{2} - b^{2} = (a + b)(a - b)\), with \(a = x^{2} + y^{2} - z^{2}\) and \(b = x^{2} - y^{2} + z^{2}\):
\[ \begin{aligned} a + b &= 2x^{2} \\ a - b &= 2y^{2} - 2z^{2} = 2(y + z)(y - z) \end{aligned} \]
\[ \begin{aligned} \frac{4x^{2}(y + z)(y - z)}{(y + z)(z - y)} &= \frac{4x^{2}(y - z)}{z - y} \\ &= -4x^{2} \end{aligned} \]
The minus sign comes from \(z - y = -(y - z)\) in the denominator; missing it gives \(4x^{2}\). The options with \(2x^{2}\) drop a factor of 2 from \(a + b\) or \(a - b\).
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