If $a+b+c=0$ find $\frac{a^2+b^2+c^2}{b^2-ca}$ I tried finding the value of $a^2+b^2+c^2$ and it is $-2(ab+bc+ca)$ but that doesn't have any common factor with $b^2-ca$ so that didn't help.
I squared on both sides of $a+b=-c$ to get $a^2+b^2-c^2+2ab=0$. But I didn't proceed any further.

With $b = \lambda a, c = \mu a$
$$
\frac{1+\lambda^2+\mu^2}{\lambda^2+\mu}=\frac{1+\lambda^2+(1+\lambda)^2}{\lambda^2+\lambda+1} = 2
$$
