Question

(x-y)(x+y)+(y-z)(y+z)+(z-x)(z+x)=0

Answer 1

Answer:

(x-y)(x+y)+(y-z)(y+z)+(z-x)(z+x)=0

x²+xy-xy-y²+y²+yz-yz-z²+z²+zx-zx-x²=0

(this all terms will cut and will get 0)

0=0

so 0=0.

Answer 2

$\huge{\boxed{\mathbb{\green{ANSWER}}}}$

$\tt{Given\:that:}$

$\sf\red{(x-y)(x+y)+(y-z)(y+z)+(z-x)(z+x)=0}$

$\sf\blue{by\:using\:the\:property:}$

$\sf\blue{(a-b)(a+b)=a^2-b^2}$

$\sf{=>(x-y)(x+y)+(y-z)(y+z)+(z-x)(z+x)=0}$

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

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

$\sf\red{so\:all\:get\:cancelled}$$\sf{therefore\:,0=0}$

$\sf{LHS=RHS}$

$\sf\pink{Hence\:proved}$

$\huge\sf\purple{Hope\:it\:helps\:you}$