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 = ?

Solution(By Examveda Team)

$$\frac{{x - {a^2}}}{{b + c}} + \frac{{x - {b^2}}}{{c + a}} + \frac{{x - {c^2}}}{{a + b}} = 4\left( {a + b + c} \right)$$
Note : In such type of question to save your valuable time assume values as per your need which make your calculation easier.
Assume a = 1, b = 0, c = 1
$$\eqalign{ & {\text{Make sure there will be no }}\left( {\frac{0}{0}} \right){\text{ form}} \cr & \therefore \frac{{x - 1}}{{1 + 0}} + \frac{{x - 0}}{{1 + 1}} + \frac{{x - 1}}{{1 + 0}} = 4 \cr & \Rightarrow x - 1 + \frac{x}{2} + x - 1 = 4 \times 2 \cr & \Rightarrow x + \frac{x}{2} + x = 8 + 2 \cr & \Rightarrow \frac{{5x}}{2} = 10 \cr & \Rightarrow x = 4 \cr & {\text{Now put values in options take option}} \cr & \left( \text{A} \right),{\left( {a + b + c} \right)^2} = {\left( {1 + 0 + 1} \right)^2} = 4 \cr} $$