Pls solve question 24!!!

Dear student

\tan x + \tan (x + \frac{π}{3}) + \tan (x + \frac{2 π}{3}) = 3 \Rightarrow \tan x + \frac{\tan x + \tan \frac{π}{3}}{1 - \tan x \tan \frac{π}{3}} + \frac{\tan x + \tan \frac{2 π}{3}}{1 - \tan x \tan \frac{2 π}{3}} = 3

\Rightarrow \tan x + \frac{\tan x + \sqrt{3}}{1 - \sqrt{3} \tan x} + \frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x} = 3

\Rightarrow \tan x + \frac{(1 + \sqrt{3} \tan x) (\tan x + \sqrt{3}) + (1 - \sqrt{3} \tan x) (\tan x - \sqrt{3})}{(1 + \sqrt{3} \tan x) (1 - \sqrt{3} \tan x)} = 3

\Rightarrow \tan x + \frac{\tan x + \sqrt{3} + \sqrt{3} \tan^{2} x + 3 \tan x + \tan x - \sqrt{3} - \sqrt{3} \tan^{2} x + 3 \tan x}{1 - 3 \tan^{2} x} = 3

\Rightarrow \tan x + \frac{8 \tan x}{1 - 3 \tan^{2} x} = 3 \Rightarrow \frac{\tan x - 3 \tan^{3} x + 8 \tan x}{1 - 3 \tan^{2} x} = 3 \Rightarrow \frac{9 \tan x - 3 \tan^{3} x}{1 - 3 \tan^{2} x} = 3 \Rightarrow \frac{3 (3 \tan x - \tan^{3} x)}{1 - 3 \tan^{2} x} = 3 \Rightarrow \frac{3 \tan x - \tan^{3} x}{1 - 3 \tan^{2} x} = 1 \Rightarrow \tan 3 x = 1

Regards