Question

Find the value of $\frac{(a+p)(a+q)}{(a-b)(a-c)(a+x)}+\frac{(b+p)(b+q)}{(b-c)(b-a)(b+x)}+\frac{(c+p)(c+q)}{(c-a)(c-b)(c+x)}$.

Answer 1

The common denominator is $(b-c)(c-a)(a-b)(x+a)(x+b)(x+c);$ the numerator

$=-\sum (a+p)(a+q)(b-c)(x+b)(x+c)$

$=-\sum (a+p)(a+q)\{x^2(b-c)+x(b^2-c^2)+bc(b-c)\}$

The co efficient of $x^2=-\sum a^2(b-c)-(p+q)\sum a(b-c)-pq\sum (b-c)$

$=(b-c)(c-a)(a-b)$

The co efficient of $x=-\sum a^2(b^2-c^2)-(p+q)\sum a(b^2-c^2)-pq\sum (b^2-c^2)$

$=(p+q)(b-c)(c-a)(a-b)$

The term independent of x

$=-abc\sum a(b-c)-abc(p+q)\sum (b-c)-pq\sum bc(b-c)$

$=pq(b-c)(c-a)(a-b)$

Thus, the numerator is $=(b-c)(c-a)(a-b)\{x^2+(p+q)x+pq\}$

Value $=\frac{x^2+(p+q)x+pq}{(x+a)(x+b)(x+c)}$