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Question

Question 5(v)
Show that:
(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0


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Solution

L.H.S = (ab)(a+b)+(bc)(b+c)+(ca)(c+a)
=a2b2+b2c2+c2a2 [Using identity (ab)(a+b)=a2b2]
=0.
=R.H.S


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