If `t a n x+tan(x+pi/3)+tan(x+(2pi)/3)=3,` then prove that `(3tanx-t a n^3x)/(1-3t a n^2x)=1` .

Explanation

To prove that if \( \tan x + \tan(x + \frac{\pi}{3}) + \tan(x + \frac{2\pi}{3}) = 3 \), then \( \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} = 1 \), we will follow these steps:

### Step 1: Simplify the left-hand side
We start with the equation:
\[ \tan x + \tan(x + \frac{\pi}{3}) + \tan(x + \frac{2\pi}{3}) = 3 \]
Using the tangent addition formula:
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