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Question

Find the value of a(ab)(ac)(xa)+b(bc)(ba)(xb)+c(ca)(cb)(xc).

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Solution

a(ab)(ac)(xa)+b(bc)(ba)(xb)+c(ca)(cb)(xc)

=a(ab)(ac)(xa)b(bc)(ab)(xb)+c(ca)(cb)(xc)

= a(bc)(xb)(xc)b(ac)(xa)(xc)+c(ab)(xa)(xb)(ab)(ac)(bc)(xa)(xb)(xc)
expanding numerator and denominator, we get

= (abac)(x2xcxb+bc)(abbc)(x2xcxa+ac)+(acbc)(x2xbxa+ab)(a2ba2cabc+ac2ab2+abc+b2cbc2)(xa)(xb)(xc)
= x(a2ba2c+ac2ab2+b2cbc2)(a2ba2c+ac2ab2+b2cbc2)(xa)(xb)(xc)

= x(xa)(xb)(xc)


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