Question : If $\frac{x - a^{2}}{b + c} + \frac{x - b^{2}}{c + a} + \frac{x - c^{2}}{a + b} = 4 (a + b + c)$ , then $x$ is equal to:
Option 1: $(a + b + c)^{2}$
Option 2: $a^{2} + b^{2} + c^{2}$
Option 3: $a b + b c + c a$
Option 4: $a^{2} + b^{2} + c^{2} - a b - b c - c a$

Question

Question : If $\frac{a^{2} - b c}{a^{2} + b c} + \frac{b^{2} - c a}{b^{2} + c a} + \frac{c^{2} - a b}{c^{2} + a b} = 1$ , then the value of $\frac{a^{2}}{a^{2} + b c} + \frac{b^{2}}{b^{2} + a c} + \frac{c^{2}}{c^{2} + a b}$ is:

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Answer 1

Correct Answer: $(a + b + c)^{2}$

Solution : Let values of $a = 1, b = 1, c = 0$
$\frac{x - a^{2}}{b + c} + \frac{x - b^{2}}{c + a} + \frac{x - c^{2}}{a + b} = 4 (a + b + c)$
Now, let us find the value of $x$ by substituting the values of $a$ , $b$ and $c$ .
$\frac{x - 1}{1} + \frac{x - 1}{1} + \frac{x}{2} = 4 (2)$
⇒ $(x - 1) + (x - 1) + \frac{x}{2} = 8$
⇒ $(2 x - 2) + \frac{x}{2} = 8$
⇒ $2 (2 x - 2) + x = 8 \times 2$
⇒ $4 x - 4 + x = 16$
⇒ $5 x = 20$
⇒ $x = 4$
Now, we will see which option gives answer 4.
The answer comes out to be $(a + b + c)^{2}$ .
Since, $(a + b + c)^{2} = 4$
Hence, the correct answer is $(a + b + c)^{2}$ .

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Question : If x−a2b+c+x−b2c+a+x−c2a+b=4(a+b+c), then x is equal to:Option 1: (a+b+c)2Option 2: a2+b2+c2Option 3: ab+bc+caOption 4: a2+b2+c2−ab−bc−ca

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Question : If $\frac{x - a^{2}}{b + c} + \frac{x - b^{2}}{c + a} + \frac{x - c^{2}}{a + b} = 4 (a + b + c)$ , then $x$ is equal to:
Option 1: $(a + b + c)^{2}$
Option 2: $a^{2} + b^{2} + c^{2}$
Option 3: $a b + b c + c a$
Option 4: $a^{2} + b^{2} + c^{2} - a b - b c - c a$