24. When $0<x<\frac{\pi}{2}$, the function $y=\tan 3 x \cdot \cot ^{3} x$ cannot take values within the open interval $(a, b)$. Find the value of $a+b$.


$$
\begin{array}{rl}
\text { 24. } \because \quad \tan 3 x=\frac{\tan x+\tan 2 x}{1-\tan x \cdot \tan 2 x}=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}, \\
\therefore y & y=\frac{3-\tan ^{2} x}{\left(1-3 \tan ^{2} x\right) \tan ^{2} x} . \\
& \text { Let } \tan ^{2} x=t \text {, then } y=\frac{3-t}{(1-3 t) t}, y(1-3 t) t=3-t,
\end{array}
$$
24. $\because \quad \tan 3 x=\frac{\tan x+\tan 2 x}{1-\tan x \cdot \tan 2 x}=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}$,

Let $\tan ^{2} x=t$, then $y=\frac{3-t}{(1-3 t) t}, y(1-3 t) t=3-t$,
which simplifies to $3 y t^{2}-(1+y) t+3=0$.
Since $t$ is a real number, the discriminant $\Delta=(1+y)^{2}-36 y=y^{2}-34 y+1 \geqslant 0$,
which means $y \leqslant 17-\sqrt{288}$ or $y \geqslant 17+\sqrt{288}$,
$$
\therefore \quad a=17-\sqrt{288}, b=17+\sqrt{288}, \therefore \quad a+b=34 .
$$