Question : Simplify the expression $\frac{s^2+t^2+2 s t-u^2}{s^2-t^2-2 t u-u^2}$, provided $(s+t+u) \neq 0$.
Option 1: $\frac{s+t-u}{s-t-u}$
Option 2: $\frac{s+t+u}{s-t+u}$
Option 3: $\frac{s-t-u}{s+t-u}$
Option 4: $\frac{s-t+u}{s+t+u}$
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Correct Answer: $\frac{s+t-u}{s-t-u}$
Solution : Given: $\frac{(s^2+t^2+2 s t)-u^2}{s^2-(t^2+2 t u+u^2)}$ We know $\small a^2+b^2 +2ab = (a+b)^2$ and $\small a^2+b^2-2ab = (a-b)^2$ So, $\frac{(s^2+t^2+2 s t)-u^2}{s^2-(t^2+2 t u+u^2)}=\frac{(s+t)^2 - u^2}{s^2 - (t+u)^2}$ $=\frac{(s+t+u)(s+t-u)}{(s+t+u)(s-t-u)}$ [using $\small a^2 - b^2 = (a+b)(a-b)$] $=\frac{s+t-u}{s-t-u}$ Hence, the correct answer is $\frac{s+t-u}{s-t-u}$.
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