Prove that the equation
\[x^3 - 3 \tan\frac{\pi}{12} x^2 - 3x + \tan\frac{\pi}{12}= 0\]
has one root $x_1 = \tan \frac{\pi}{36}$, and find the other roots.
1. **Verify that \( x_1 = \tan \frac{\pi}{36} \) is a root of the given equation:**

   We start with the given equation:
   \[
   x^3 - 3 \tan\frac{\pi}{12} x^2 - 3x + \tan\frac{\pi}{12} = 0
   \]
   Substitute \( x_1 = \tan \frac{\pi}{36} \) into the equation:
   \[
   \left( \tan \frac{\pi}{36} \right)^3 - 3 \tan\frac{\pi}{12} \left( \tan \frac{\pi}{36} \right)^2 - 3 \tan \frac{\pi}{36} + \tan\frac{\pi}{12} = 0
   \]
   To simplify, we use the triple angle identity for tangent:
   \[
   \tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}
   \]
   Let \( x = \frac{\pi}{36} \), then \( 3x = \frac{\pi}{12} \). Therefore:
   \[
   \tan \frac{\pi}{12} = \frac{3 \tan \frac{\pi}{36} - \left( \tan \frac{\pi}{36} \right)^3}{1 - 3 \left( \tan \frac{\pi}{36} \right)^2}
   \]
   Rearrange this equation to match the form of the given polynomial:
   \[
   \left( \tan \frac{\pi}{36} \right)^3 - 3 \tan\frac{\pi}{12} \left( \tan \frac{\pi}{36} \right)^2 - 3 \tan \frac{\pi}{36} + \tan\frac{\pi}{12} = 0
   \]
   This confirms that \( x_1 = \tan \frac{\pi}{36} \) is indeed a root.

2. **Find the other roots:**

   We know that the polynomial can be factored using the roots derived from the triple angle identity. The roots of the polynomial \( x^3 - 3 \tan\frac{\pi}{12} x^2 - 3x + \tan\frac{\pi}{12} = 0 \) are given by:
   \[
   \tan u, \tan \left( u + \frac{\pi}{3} \right), \tan \left( u + \frac{2\pi}{3} \right)
   \]
   where \( u = \frac{\pi}{36} \). Therefore, the roots are:
   \[
   \tan \frac{\pi}{36}, \tan \left( \frac{\pi}{36} + \frac{\pi}{3} \right), \tan \left( \frac{\pi}{36} + \frac{2\pi}{3} \right)
   \]
   Simplify the arguments of the tangent functions:
   \[
   \tan \left( \frac{\pi}{36} + \frac{\pi}{3} \right) = \tan \left( \frac{\pi}{36} + \frac{12\pi}{36} \right) = \tan \frac{13\pi}{36}
   \]
   \[
   \tan \left( \frac{\pi}{36} + \frac{2\pi}{3} \right) = \tan \left( \frac{\pi}{36} + \frac{24\pi}{36} \right) = \tan \frac{25\pi}{36}
   \]
   Thus, the roots of the polynomial are:
   \[
   \boxed{\left\{ \tan \frac{\pi}{36}, \tan \frac{13\pi}{36}, \tan \frac{25\pi}{36} \right\}}
   \]