226. Equation containing fractions. Solve the equation

$$
\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b}
$$
226. By direct substitution, we can verify that $x=0$ and $x=a+b$ are solutions to our equation. Moreover, if $m+n=\frac{1}{m}+\frac{1}{n}$, then $(m+n) \times(m n-1)=0$. Therefore, for $x \neq 0$ and $\neq a+b$

$$
\frac{x-a}{b}+\frac{x-b}{a}=0, \quad \text { that is } \quad(a+b) x=a^{2}+b^{2}
$$

Thus, the third root of our equation will be $\frac{\left(a^{2}+b^{2}\right)}{(a+b)}$.