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Question

Evaluate (a+b)(ab)+(b+c)(bc)+(c+a)(ca).

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Solution

We are given,
(a+b)(ab)+(b+c)(bc)+(c+a)(ca)
=(a2b2)+(b2c2)+(c2a2)
=(a2a2)+(b2b2)+(c2c2)
=0
So, (a+b)(ab)+(b+c)(bc)+(c+z)(ca)

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