Question

If $a,b,c\in R$, then prove that the roots of the equation $\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}=0$ are always real and cannot have roots if $a=b=c$.

Answer 1

$\frac{1}{(x-a)}+\frac{1}{(x-b)}+\frac{1}{(x-c)}=0$

$(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0$

$x^2-bx-cx+bc+x^2-ax-cx+ac+x^2-bx-ax+ab=0$

$3x^2-2(a+b+c)x+(ab+bc+ac)=0$

$D=b^2-4ac$

$=4(a+b+c{)}^2-4\times 3\times (ab+bc+ac)$

$=4(a^2+b^2+c^2+2(ab+bc+ac)-3(ab+bc+ac\left)\right)$

$=4(a^2+b^2+c^2-(ab+bc+ac\left)\right)$

Which is real ,

Also the equation does not exist if $a=b=c$.