If a + b + c = 0, find the value of $$\frac{{{a^2}}}{{\left( {{a^2} - bc} \right)}} + \frac{{{b^2}}}{{\left( {{b^2} - ca} \right)}} + \frac{{{c^2}}}{{\left( {{c^2} - ab} \right)}} = ?$$

Explanation:-

$$\eqalign{ & a + b + c = 0 \cr & \Rightarrow a = - \left( {b - c} \right) \cr & \Rightarrow {a^2} = {\left( {b + c} \right)^2} \cr & \therefore \frac{{{a^2}}}{{\left( {{a^2} - bc} \right)}} + \frac{{{b^2}}}{{\left( {{b^2} - ca} \right)}} + \frac{{{c^2}}}{{\left( {{c^2} - ab} \right)}} \cr & = \frac{{{{\left( {b + c} \right)}^2}}}{{{{(b + c)}^2} - bc}} + \frac{{{b^2}}}{{{b^2} + c\left( {b + c} \right)}} + \frac{{{c^2}}}{{{c^2} + b\left( {b + c} \right)}} \cr & = \frac{{{{\left( {b + c} \right)}^2}}}{{{b^2} + {c^2} + bc}} + \frac{{{b^2}}}{{{b^2} + {c^2} + bc}} + \frac{{{c^2}}}{{{b^2} + {c^2} + bc}} \cr & = \frac{{{b^2} + {c^2} + 2bc + {b^2} + {c^2}}}{{{b^2} + {c^2} + bc}} \cr & = \frac{{2\left( {{b^2} + {c^2} + bc} \right)}}{{{b^2} + {c^2} + bc}} \cr & = 2 \cr}$$