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

Solution : Given: $\frac{x+a^2+2c^2}{b+c}+\frac{x+b^2+2a^2}{c+a}+\frac{x+c^2+2b^2}{a+b}=0$
We know the algebraic identity, $(x^2-y^2)=(x+y)(x-y)$.
Substitute the value of $x=-(a^2+b^2+c^2)$ in the given equation, we get,
$\frac{–a^2–b^2–c^2+a^2+2c^2}{b+c}+\frac{–a^2–b^2–c^2+b^2+2a^2}{c+a}+\frac{–a^2–b^2–c^2+c^2+2b^2}{a+b}=0$
$⇒\frac{c^2–b^2}{b+c}+\frac{a^2–c^2}{c+a}+\frac{b^2–a^2}{a+b}=0$
$⇒\frac{(c–b)(b+c)}{b+c}+\frac{(a–c)(c+a)}{c+a}+\frac{(b–a)(a+b)}{a+b}=0$
$⇒c-b+a-c+b-a=0$
$⇒0=0$
It satisfies the equation.
Hence, the correct answer is $-(a^2+b^2+c^2)$.