Question : If $\frac{a^{2}}{b + c} = \frac{b^{2}}{c + a} = \frac{c^{2}}{a + b} = 1$ , then find the value of $\frac{2}{1 + a} + \frac{2}{1 + b} + \frac{2}{1 + c}$ .
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3

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Question : If $a + b + c = 0$ , then the value of $\frac{1}{(a + b) (b + c)} + \frac{1}{(b + c) (c + a)} + \frac{1}{(c + a) (a + b)}$ is:

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Answer 1

Correct Answer: 2

Solution : Given: $\frac{a^{2}}{b + c} = \frac{b^{2}}{c + a} = \frac{c^{2}}{a + b} = 1$
⇒ $a^{2} = b + c$ , $b^{2} = c + a$ , and $c^{2} = a + b$
Now, $\frac{2}{1 + a} + \frac{2}{1 + b} + \frac{2}{1 + c}$
Multiplying the numerator and denominator of the first, second, and third term by $a$ , $b$ , and $c$ respectively.
$= \frac{2 a}{a + a^{2}} + \frac{2 b}{b + b^{2}} + \frac{2 c}{c + c^{2}}$
$= \frac{2 a}{a + b + c} + \frac{2 b}{b + a + c} + \frac{2 c}{c + a + b}$
$= \frac{2 a + 2 b + 2 c}{a + b + c}$
$= \frac{2 (a + b + c)}{a + b + c}$
$= 2$
Hence, the correct answer is 2.

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Question : If a2b+c=b2c+a=c2a+b=1, then find the value of 21+a+21+b+21+c.Option 1: 0Option 2: 1Option 3: 2Option 4: 3

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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 $\frac{2}{1 + a} + \frac{2}{1 + b} + \frac{2}{1 + c}$ .
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3