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...
$$\frac{1}{a^2+ab+ca}+\frac{1}{b^2+ab+bc}+\frac{1}{c^2+ca+bc}$$
$$=\frac{1}{a(a+b+c)}+\frac{1}{b(a+b+c)}+\frac{1}{c(a+b+c)}=\frac{ab+bc+ca}{abc(a+b+c)}=0.$$
$ab+bc+ca=0$

$a(b+c)= -bc$

$b+c= -bc/a$

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

$1/a(a+b+c)+ 1/b(a+b+c)+ 1/c(a+b+c)$