Question

Show that (x-y) (x-y)+(y-z)(y+z) + (z-x)(z+x) = 0​

Answer 1

$\rm{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}$

$\rm{=x^2-y^2+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}$

$\rm{=x^2-y^2+y^2-z^2+\left(z-x\right)\left(z+x\right)}$

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

Answer 2

Step-by-step explanation:

(x²-y²) +(y²-z²) +(z²-x²)

x²-y²+y²-z²+z²-x²

=0