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.

It's simple algebra to verify$$\frac{a^2+b^2+(a+b)^2}{b^2+a(a+b)}=2$$provided the denominator is nonzero, but it's $(a+b/2)^2+3b^2/4$, so that only requires $a,\,b$ to not both be $0$.
