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Question

Find the value of a2b2c2(ab)(ac)+b2c2a2(bc)(ba)+c2a2b2(ca)(cb).

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Solution

a2b2c2(ab)(ac)+b2c2a2(bc)(ba)+ c2a2b2(ca)(cb)

=a2b2c2(ab)(ac) b2c2a2(bc)(ab) + c2a2b2(ca)(cb)


=(a2b2c2)(bc)(b2c2a2)(ac)+(c2a2b2)(ab)(ab)(ac)(bc)
Expanding numerator and denominator we get,

= a2ba2cb3+b2cc2b+c3ab2+b2c+ac2c3+a3a2c+ac2bc2a3+a2b+b3ab2a2ba2cabc+ac2ab2+abc+b2cbc2


= 2(a2ba2c+ac2ab2+b2cbc2)a2ba2c+ac2ab2+b2cbc2

=2

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