If xy + yz + zx = 0, then $$\left( {\frac{1}{{{x^2} - yz}} + \frac{1}{{{y^2} - zx}} + \frac{1}{{{z^2} - xy}}} \right)$$       $$\left( {x,y,z \ne 0} \right) = ?$$

Solution(By Examveda Team)

$$\eqalign{ & xy + yz + zx = 0 \cr & \therefore xy + zx = - yz \cr & \Rightarrow xy + yz = - zx \cr & \Rightarrow yz + zx = - xy \cr & \therefore \frac{1}{{{x^2} - yz}} + \frac{1}{{{y^2} - zx}} + \frac{1}{{{z^2} - xy}} \cr} $$
Putting values of -yz, -zx, -xy from above
$$ \Rightarrow \frac{1}{{{x^2} + \left( {xy + zx} \right)}} + \frac{1}{{{y^2} + \left( {xy + yz} \right)}}$$       $$ + \frac{1}{{{z^2} + \left( {yz + zx} \right)}}$$
$$ \Rightarrow \frac{1}{{x\left( {x + y + z} \right)}} + \frac{1}{{y\left( {x + y + z} \right)}}$$       $$ + \frac{1}{{z\left( {x + y + z} \right)}}$$
$$\eqalign{ & \Rightarrow \frac{1}{{\left( {x + y + z} \right)}}\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) \cr & \Rightarrow \frac{1}{{\left( {x + y + z} \right)}}\left( {\frac{{zy + xz + xy}}{{xyz}}} \right) \cr & \Rightarrow \frac{1}{{x + y + z}} \times 0 \cr & \Rightarrow 0 \cr} $$