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: $ab+bc+ca$
Option 4: $a^{2}+b^{2}+c^{2}-ab-bc-ca$
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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 $c$.
$\frac{x-1}{1}+\frac{x-1}{1}+\frac{x}{2}=4(2)$
⇒ $(x-1)+(x-1)+\frac{x}{2}=8$
⇒ $(2x-2)+\frac{x}{2}=8$
⇒ $2(2x-2)+x=8\times 2$
⇒ $4x-4+x=16$
⇒ $5x=20$
⇒ $x=4$
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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