Solve the following quadratic equations by factorization:

We have been given

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

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

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

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

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

Therefore,

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

$x=0$

or,

$x-(a+b)=0$

$x=a+b$

Hence, $x=0$ or $x=a+b$.