<s>[INST] 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.
 [/INST] $0 = 0^2 = (a + b + c)^2 = a^2 + b^2 + c^2  + 2ab + 2ac + 2bc$
$a^2 + b^2 + c^2 = -2(ab + ac + bc)$
$b² - ac = -b(a + c) - ac = -(ab + ac + bc)$  as  $b = -(a + c)$
Hence, the answer is 2.
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