<s>[INST] If $ab+bc+ca=0$ then the value of $1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is If $ab+bc+ca=0$
then the value of 
$1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is...
 [/INST] $ab+bc+ca=0$
$a(b+c)= -bc$
$b+c= -bc/a$
Adding both side $a$ then $b+c+a=a-bc/a$
$a(a+b+c)= a^2-bc$
$1/(a^2-bc)= 1/a(a+b+c)$
Similarly $1/(b^2-ca) = 1/b(a+b+c)$
and $1/(c^2-ab) = 1/c(a+b+c)$.
By adding all eqution we get
$1/a(a+b+c)+ 1/b(a+b+c)+ 1/c(a+b+c)$
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