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Question

Find the value of (a+p)(a+q)(ab)(ac)(a+x)+(b+p)(b+q)(bc)(ba)(b+x)+(c+p)(c+q)(ca)(cb)(c+x).

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Solution

The common denominator is (bc)(ca)(ab)(x+a)(x+b)(x+c); the numerator
=(a+p)(a+q)(bc)(x+b)(x+c)
=(a+p)(a+q){x2(bc)+x(b2c2)+bc(bc)}
The co efficient of x2=a2(bc)(p+q)a(bc)pq(bc)
=(bc)(ca)(ab)
The co efficient of x=a2(b2c2)(p+q)a(b2c2)pq(b2c2)
=(p+q)(bc)(ca)(ab)
The term independent of x
=abca(bc)abc(p+q)(bc)pqbc(bc)
=pq(bc)(ca)(ab)
Thus, the numerator is =(bc)(ca)(ab){x2+(p+q)x+pq}
Value =x2+(p+q)x+pq(x+a)(x+b)(x+c)

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