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

Solution (By Examveda Team)

$$\eqalign{ & {\text{Given xepression ,}} \cr & \frac{{\left( {x + y - z} \right)\left( {x - y + z} \right)}}{{\left( {x + y + z} \right)\left( {x + z - y} \right)}} + \frac{{\left( {y + x - z} \right)\left( {y - x + z} \right)}}{{\left( {x + y + z} \right)\left( {x + y - z} \right)}} + \frac{{\left( {z + x - y} \right)\left( {z - x + y} \right)}}{{\left( {y + z + x} \right)\left( {y + z - x} \right)}} \cr & = \frac{{\left( {x + y - z} \right)}}{{\left( {x + y + z} \right)}} + \frac{{\left( {y - x + z} \right)}}{{\left( {x + y + z} \right)}} + \frac{{\left( {x - y + z} \right)}}{{\left( {x + y + z} \right)}} \cr & = \frac{{\left( {x + y - z} \right) + \left( {y - x + z} \right) + \left( {x - y + z} \right)}}{{\left( {x + y + z} \right)}} \cr & = \frac{{x + y + z}}{{x + y + z}} \cr & = 1 \cr} $$