The value of $$\frac{{{{\left( {x - y} \right)}^3} + {{\left( {y - z} \right)}^3} + {{\left( {z - x} \right)}^3}}}{{{{\left( {{x^2} - {y^2}} \right)}^3} + {{\left( {{y^2} - {z^2}} \right)}^3} + {{\left( {{z^2} - {x^2}} \right)}^3}}}$$       is = ?

Solution (By Examveda Team)

$$\eqalign{ & {\text{Since }}\left( {x - y} \right) + \left( {y - z} \right) + \left( {z - x} \right) = 0 \cr & {\text{So,}}{\left( {x - y} \right)^3} + {\left( {y - z} \right)^3} + {\left( {z - x} \right)^3} \cr & = 3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right) \cr & {\text{Since}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) = 0 \cr & {\text{So,}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) \cr & = 3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right) \cr & \therefore {\text{Given expression }} \cr & {\text{ = }}\frac{{3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}}{{3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}} \cr & = \frac{1}{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}} \cr & = {\left[ {\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)} \right]^{ - 1}} \cr} $$