If ab + bc+ ca = 0, then the value of $$\frac{1}{a^2-bc}+\frac{1}{b^2-ac}+\frac{1}{c^2-ab}$$

Since, $$ab + bc + ca = 0$$

If we take $$a = b = 1$$, => $$c = \frac{-1}{2}$$

To find : $$\frac{1}{a^2-bc}+\frac{1}{b^2-ac}+\frac{1}{c^2-ab}$$

Substituting values of $$a,b,c$$ we get :

= $$\frac{1}{1 - (\frac{-1}{2})} + \frac{1}{1 - (\frac{-1}{2})} + \frac{1}{(\frac{1}{4}) - 1}$$

= $$\frac{1}{\frac{3}{2}} + \frac{1}{\frac{3}{2}} + \frac{1}{\frac{-3}{4}}$$

= $$\frac{2}{3} + \frac{2}{3} - \frac{4}{3}$$

= $$\frac{4}{3} - \frac{4}{3} = 0$$