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
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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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Question : If $\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1$, then find the value of $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$.
Question : If $x+\frac{1}{x}=c+\frac{1}{c}$, then the value of $x$ is:
Question : If $\frac{2+a}{a}+\frac{2+b}{b}+\frac{2+c}{c}=4$, then the value of $\frac{ab+bc+ca}{abc}$ is:
Question : If $a+b+c=0$, the value of $\frac{a^2+b^2+c^2}{a^2-bc}$ is:
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