Question

..
The value of (x-y) (x+y)+ (y-z)
(y+z) + (z-x) (z+x) is equal to:
(a) 0
(b) 1
(c) 2
(d) xyz​

Answer 1

Answer:

The answer is option (A)0

(x-y)(x+y)+(y-z)(y+z)+(z-x)(z+x)

$= > {x }^{2} - {y}^{2} + {y}^{2} - {z }^{2} + {z}^{2} - {x}^{2} \\ = > 0$