Question : If $\frac{a^{2}}{b + c} = \frac{b^{2}}{c + a} = \frac{c^{2}}{a + b} = 1$ , then $\frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c}$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 4

Question

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: 1

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

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Question : If a2b+c=b2c+a=c2a+b=1, then 11+a+11+b+11+c is:Option 1: 1Option 2: 2Option 3: 3Option 4: 4

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Question : If $\frac{a^{2}}{b + c} = \frac{b^{2}}{c + a} = \frac{c^{2}}{a + b} = 1$ , then $\frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c}$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 4