Question : The value of $\frac{a^2-(b-c)^2}{(a+c)^2-b^2}+\frac{b^2-(a-c)^2}{(a+b)^2-c^2}+\frac{c^2-(a-b)^2}{(b+c)^2-a^2}$ is:
Option 1: 1
Option 2: 2
Option 3: –1
Option 4: 3
Correct Answer: 1
Solution :
$\frac{a^2-(b-c)^2}{(a+c)^2-b^2}+\frac{b^2-(a-c)^2}{(a+b)^2-c^2}+\frac{c^2-(a-b)^2}{(b+c)^2-a^2}$
= $\frac{(a-b+c)(a+b-c)}{(a+c-b)(a+c+b)} + \frac{(b-a+c)(b+a-c)}{(a+b-c)(a+b+c)} + \frac{(c-a+b)(c+a-b)}{(b+c-a)(b+c+a)}$
= $\frac{(a+b-c)}{(a+c+b)} + \frac{(b-a+c)}{(a+b+c)} + \frac{(c+a-b)}{(b+c+a)}$
= $\frac{a+b-c+b-a+c+c+a-b}{a+b+c}$
= $\frac{a+b+c}{a+b+c}$
= $1$
Hence, the correct answer is 1.
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