The roots of x , if x−aba+b+x−aca+c+x−bcb+c=a+b+c

x−aba+b+x−aca+c+x−bcb+c=a+b+cx−aba+b−c+x−bcb+c−a+x−cac+a−b=0x−ab−ac−bca+b+x−ab−ac−bcb+c+x−ab−ac−bcc+a=0(x−ab−ac−bc)1a+b+1b+c+1c+a=0(x−ab−ac−bc)=0x=(ab+bc+ca)

The roots of x , if x−aba+b+x−aca+c+x−bcb+c=a+b+c

x−aba+b+x−aca+c+x−bcb+c=a+b+cx−aba+b−c+x−bcb+c−a+x−cac+a−b=0x−ab−ac−bca+b+x−ab−ac−bcb+c+x−ab−ac−bcc+a=0(x−ab−ac−bc)1a+b+1b+c+1c+a=0(x−ab−ac−bc)=0x=(ab+bc+ca)

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The roots of x , if $\frac{x - a b}{a + b} + \frac{x - a c}{a + c} + \frac{x - b c}{b + c} = a + b + c$

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$x = (a b + b c + c a)$

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

answer is B.

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Detailed Solution

$\frac{x - a b}{a + b} + \frac{x - a c}{a + c} + \frac{x - b c}{b + c} = a + b + c$

$\begin{array}{l} \frac{x - a b}{a + b} - c + \frac{x - b c}{b + c} - a + \frac{x - c a}{c + a} - b = 0 \\ \begin{array}{ll} \frac{x - a b - a c - b c}{a + b} & + \frac{x - a b - a c - b c}{b + c} + \frac{x - a b - a c - b c}{c + a} = 0 \\ (x - a b - a c - b c) & (\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a}) = 0 \\ (x - a b - a c - b c) = & 0 \\ x = (a b + b c + c a) \end{array} \end{array}$