Q25 of 50 Page 9

Show that (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

L.H.S = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)


Using identity:


(a – b) (a + b) = a2 – b2


We get,


= (a2 – b2) + (b2 – c2) + (c2 – a2)


= a2 – b2 + b2 – c2 + c2 – a2


= 0


= R.H.S


Hence, verified


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