Question : If $a+b+c = 0$, then the value of $\small \frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}$ is:
Option 1: 0
Option 2: 1
Option 3: 3
Option 4: 2
Correct Answer: 0
Solution :
Given: $a+b+c = 0$
Solution:
$\frac{1}{(a+b)(b+c)}+\frac{1}{(b+c)(c+a)}+\frac{1}{(c+a)(a+b)}$
Taking the LCM, we get
= $\frac{(c+a)+(a+b)+(b+c)}{(a+b)(b+c)(c+a)}$
= $\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}$
Substituting the value of $(a+b+c)$ in the equation:
= $\frac{2\times 0}{(a+b)(b+c)(c+a)}$
= 0
Hence, the correct answer is 0.
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