A problem by rana kar

$ab+bc+ca=0$ implies $-ab=bc+ca$ , $-bc=ab+ca$ and $-ca=ab+bc$ . So the expression above $\frac{1}{a^2-bc}+\frac{1}{b^2-ca}+\frac{1}{c^2-ab}$ becomes $\frac{1}{a^2+(ab+ca)}+\frac{1}{b^2+(ab+bc)}+\frac{1}{c^2+(bc+ca)}.$ Notice that the common factor of the three terms is $\frac{1}{a+b+c}$ . So, factoring this out and then combining terms gives $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\frac{1}{a+b+c}=\frac{ab+ac+bc}{abc} \cdot \frac{1}{a+b+c}.$ Since $ab+ac+bc=0$ , the entire expression zeroes out. Thus, the answer is $\boxed{0}$ .

Given: ab + bc + ca = 0

Putting -bc = ab + ca, -ca = ab + bc, and -ab = bc + ca, we get the following:

a^2 - bc = a^2 + ab + ca = a(a + b + c)

b^2 - ca = b^2 + ab + bc = b(a + b + c)

c^2 - ab = c^2 + bc + ca = c(a + b + c)

Therefore, the given expression simplifies into:

1/{a(a + b + c)} + 1/{b(a + b + c)} + 1/{c(a + b + c)}

= (bc + ca + ab)/{abc(a + b + c}

= 0/{abc(a + b + c}

= 0