Question

$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0$

Answer 1

$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0$

$LHS=a^2-b^2-ab+ab+b^2-c^2+bc-bc+c^2-a^2-ac+ac$

we see the terms cancel each other hence equation reduced to zero

Or we can also use identity $(a-b)(a+b)=a^2-b^2$ to find similar result.