$\mathrm{If} \mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }, \mathrm{\delta }$ satisfy the equation $\tan \left[\mathrm{x}+\frac{\mathrm{\pi }}{4}\right]=3 \tan 3\mathrm{x} \mathrm{then} \tan \mathrm{\alpha } \tan \mathrm{\beta } \tan \mathrm{\gamma } \tan \mathrm{\delta }=$

$$\begin{gathered} \tan \left[\mathrm{x}+\frac{\mathrm{\pi }}{4}\right]=3 \tan 3\mathrm{x} \\ \frac{1+\tan x}{1-\tan x}=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x} \\ \left(1+\tan x\right)\left(1-3{\tan }^{2}x\right)=3\left(1-\tan x\right)\left(3\tan x-{\tan }^{3}x\right) \\ 1+\tan x-3{\tan }^{2}x-3{\tan }^{3}x=9\tan x-9{\tan }^{2}x-3{\tan }^{3}x+3{\tan }^{4}x \\ 3{\tan }^{4}x-6{\tan }^{2}x+8\tan x-1=0 \\ \tan \alpha \tan \beta \tan \gamma \tan \delta =-\frac{1}{3} \end{gathered}$$