Question

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0​

Answer 1

LHS

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)

=a^2-b^2+b^2-c^2+c^2-a^2

=0=RHS

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Answer 2

>> Yes, It will be 0, as. -

(a-b)(a+b)=a^2-b^2

(b-c)(b+c)=b^2-c^2

(c-a)(c+a)=c^2-a^2

So,

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

>> a^2-b^2+b^2-c^2+c^2-a^2

Which is when solved,gives 0.

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