If $\alpha =2\pi/7$, find $\tan\alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha$. 
If $\alpha =2\pi/7$, find $\tan\alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha$.

I tried by $7\alpha = 2\pi$
$4\alpha =2\pi-3\alpha$
$\sin 4\alpha = \sin 3\alpha$
But couldn't reach solution Please help

$\displaystyle \tan \frac{k \pi}{7}; 1 \le k \le 6$ are roots of 
$\tan 7 \theta = 0$ which gives the polynomial $$y^6 - \binom{7}{2} y^4 + \binom{7}{4} y^2- 7=0$$ with the substitution $ y = \tan \theta$
This also yields that $\displaystyle \tan^2 \frac{k \pi}{7}; k={2,4,8}$ are roots to the polynomial $$y^3 - \binom{7}{2} y^2 + \binom{7}{4} y - 7=0$$ so that from Vieta's formula $$\tan^2 \frac{2 \pi}{7}+\tan^2 \frac{4 \pi}{7}+\tan^2 \frac{8 \pi}{7} = 21$$
Further we have $$\tan \frac{2 \pi}{7}+\tan \frac{4 \pi}{7}+\tan \frac{8 \pi}{7} =  \tan \frac{2 \pi}{7}\tan \frac{4 \pi}{7}\tan \frac{8 \pi}{7}$$
But from Vieta, $$tan^2 \frac{2 \pi}{7} \tan^2 \frac{4 \pi}{7} \tan^2 \frac{8 \pi}{7} = 7$$ and hence $$\tan \frac{2 \pi}{7}\tan \frac{4 \pi}{7}\tan \frac{8 \pi}{7} = -\sqrt 7$$
Using the identity 2$$\sum ab =(a+b+c)^2 -(a^2+b^2+c^2)$$ we obtain the required sum as $-7$
