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

Answers (1)

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Opening all brackets from the LHS, we get : 

                 (a-b)(a+b) + (b-c) ( b +c)+(c-a)(c+a)\\\\ =\ a^2 +ab - ab- b^2 + b^2+bc - bc -c^2 + c^2 +ca - ac -a^2

                 = 0              =  RHS

Posted by

Devendra Khairwa

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