Solve the following quadratic equations by factorization:

We have been given,

$\frac{a}{(x-a)}+\frac{b}{(x-b)}=\frac{2 c}{(x-c)}$

$a(x-b)(x-c)+b(x-a)(x-c)=2 c(x-a)(x-b)$

$a\left(x^{2}-(b+c) x+b c\right)+b\left(x^{2}-(a+c) x+a c\right)=2 c\left(x^{2}-(a+b) x+a b\right)$

$(a+b-2 c) x^{2}-(2 a b-a c-b c) x=0$

$x[(a+b-2 c) x-(2 a b-a c-b c)]=0$

Therefore,

$x=0$

or,

$(a+b-2 c) x-(2 a b-a c-b c)=0$

$(a+b-2 c) x=(2 a b-a c-b c)$

$x=\frac{(2 a b-a c-b c)}{(a+b-2 c)}$

Hence, $x=0$ or $x=\frac{(2 a b-a c-b c)}{(a+b-2 c)}$.