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Question

Prove (2a2b)a+b×(2b2c)b+c×(2c2a)c+a=1

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Solution

(2a2b)a+b×(2b2c)b+c×(2c2a)c+a=(2ab)a+b+(2bc)b+c+(2ca)c+a (since, aman=amn)

=2(ab)(a+b)×2(bc)(b+c)×2(ca)(c+a) (since, (am)n=amn)

=2(a2b2)×2(b2c2)×2(c2a2)

=2a2b2+b2c2+c2a2 (since, am×an=am+n)

=20=1

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