Find the value of $m$ if $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}=3$ 
If $\dfrac{m-a^{2}}{b^{2}+c^{2}}+\dfrac{m-b^{2}}{a^{2}+c^{2}}+\dfrac{m-c^{2}}{b^{2}+a^{2}}=3;\ \ m,a,b,c \in\mathbb{R}$
Then the value of $m$ is...
Options 
$\boldsymbol{1.)}\ a^{2}-b^{2}-c^{2} \quad  \quad
\boldsymbol{2.)}\ a^{2}+b^{2}-c^{2}\\
\boldsymbol{3.)}\ a^{2}+b^{2} \quad  \quad   \quad   \quad
\boldsymbol{4.)}\ a^{2}+b^{2}+c^{2}$

By observation if all the three value 
$\dfrac{m-a^{2}}{b^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=1$
Then $m=a^{2}+b^{2}+c^{2}$
I want to know if their is any other simple and short method other than the stated observation .
I have studied maths up to $12$th grade.

the given equation is equivalent to $$-\left(a^4+3 a^2 b^2+3 a^2 c^2+b^4+3 b^2 c^2+c^4\right)
   \left(a^2+b^2+c^2-m\right)=0$$ thus we obtain $$m=a^2+b^2+c^2$$
