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