19. For $0<x<\frac{\pi}{2}$, let $\frac{\operatorname{ctg}^{3} x}{\operatorname{ctg} 3 x}$ take all real numbers except those in the open interval $(a, b)$. Find $a+b$.

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Note: The function $\operatorname{ctg} x$ is the cotangent function, often denoted as $\cot x$ in English.
19. Notice that $\operatorname{ctg} 3 x=\frac{\operatorname{ctg}^{3} x-3 \operatorname{ctg} x}{3 \operatorname{ctg}^{2} x-1}$. Let $y=\frac{\operatorname{ctg}^{3} x}{\operatorname{ctg} 3 x}, t=\operatorname{ctg}^{2} x$, then
$$
3 t^{2}-(y+1)t+3 y=0 \text {. }
$$

Thus this equation has real roots, hence
$$
\Delta=(y+1)^{2}-36 y=y^{2}-34 y+1
$$

must be non-negative. And the roots of $y^{2}-34 y+1=0$ are
$$
a=17-12 \sqrt{2} \text { and } b=17+12 \sqrt{2} \text {. }
$$

Therefore, $\Delta \geqslant 0$ if and only if $y \leqslant a$ or $y \geqslant b$. Every $y$ value outside of $(a, b)$ is attainable, and thus, we have
$$
a+b=34 \text {. }
$$