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.

Let $a^2=x\;,b^2=y\;,c^2=z\;,$ Then equation convert into $\displaystyle \frac{m-x}{y+z}+\frac{m-y}{z+x}+\frac{m-z}{x+y}=1+1+1$
So $\displaystyle \left(\frac{m-x}{y+z}-1\right)+\left(\frac{m-y}{z+x}-1\right)+\left(\frac{m-z}{x+y}-1\right)=0$
So $\displaystyle (m-x-y-z)\left\{\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right\}=0$
So  either $m=x+y+z$ or $\displaystyle \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}=0$
So $m=x+y+z = a^2+b^2+c^2$ 
