If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$ If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$ 
My 1st  approach : 
$\tan(\alpha +2\alpha +4\alpha) = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $
$\Rightarrow 0 = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $ which doesn't give me any solution. 
My IInd approach : 
U\sing Euler substitution : 
\since $\cos\theta +i\sin\theta = e^{i\theta} $.....(i) and $\cos\theta -i\sin\theta =e^{-i\sin\theta}$....(ii)
Adding (i) and (ii) we get $\cos\theta =\frac{e^{i\theta} +e^{-i\theta}}{2}$  and subtracting (i) and (ii) we get $\sin\theta =\frac{e^{i\theta} -e^{-i\theta}}{2}$ 
By u\sing this we can write : $$\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$$ as $$\frac{1}{4}\left[ (e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}) (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}}) + (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}})(e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) + (e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) (e^{\frac{i\pi}{7}} -e^{\frac{-i\pi}{7}})\right]$$
$$\large= e^{i\frac{6\pi}{7}}-e^{\frac{i2\pi}{7}}-e^{\frac{-i2\pi}{7}} +e^{\frac{-i6\pi}{7}} +e^{\frac{i3\pi}{7}}-e^{\frac{-i5\pi}{7}}-e^{\frac{i5\pi}{7}} +e^{\frac{-3\pi}{7}} +e^0 -e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}+e^0$$
Can anybody please suggest whether this is my correct approach or not. please guide further... Thanks.

We can usually solve these kind of problems by building an equation for which the the given trigonometric values are the roots. (See S.L.Loney's book)
It's as much time consuming as doing it by expanding and simplifying the angle sums, but we get several other answer's easily once the equation is in place.
Here is an overview of the method:
First, use (can be derived from Euler's formula):
\begin{align*}
  \cos{(7\, \alpha)} &= 64\cos(\alpha)^7-112\cos(\alpha)^5+56\cos(\alpha)^3-7\cos(\alpha)
\end{align*}
Substituting for $\alpha$ and factoring gives:
\begin{align*}
  8c^3+4c^2-4c-1 &= 0 \tag 1
\end{align*}
which is the equation whose roots are $\cos\left(\frac{2\pi}{7}\right),\cos\left(\frac{4\pi}{7}\right)$ and $\cos\left(\frac{2\pi}{7}\right)$
Substituting $c=\frac{1}{c}$ in $(1)$ gives:
\begin{align*}
  c^3+4c^2-4c-8 &= 0 \tag 2
\end{align*}
which has the roots $\sec \left(\frac{2\pi}{7}\right),\sec\left(\frac{4\pi}{7}\right)$ and $\sec\left(\frac{2\pi}{7}\right)$
Substituting $c=\sqrt{1+c}$ in $(2)$ and rationalizing gives:
\begin{align*}
  c^3-21c^2+35c-7 &= 0 \tag 3
\end{align*}
which has the roots $\tan \left(\frac{2\pi}{7}\right)^2 ,\tan\left(\frac{4\pi}{7}\right)^2$ and $\tan\left(\frac{2\pi}{7}\right)^2$
Getting further equations where roots will be square of the current roots is easy, but going back to find square roots is a bit ambiguous for me, I'd like to know if there 
is a better way. I did it like this:
Consider an equation, 
\begin{align*}
  x^3-a\, x^2+b\, x - c &= 0 \tag 4
\end{align*}
Substitute $x=\sqrt{x}$ and rationalize:
\begin{align*}
  x^3-(a^2 - 2\, b)x^2+(b^2 - 2ac)\, x - c^2 &= 0 \tag 5
\end{align*}
Compare with the equation $(2)$ (subst. $c=x$ there)
\begin{align*}
  x^3-21x^2+35x-7 &= 0 \tag 6
\end{align*}
and solve for $a,b,c$ and choose by inspection:
$(4)$ would be:
\begin{align*}
  x^3+\sqrt{7}x^2-7x+\sqrt{7} &= 0 \tag 7
\end{align*}
which is the equation having $\tan \left(\frac{2\pi}{7}\right) ,\tan\left(\frac{4\pi}{7}\right)$ and $\tan\left(\frac{2\pi}{7}\right)$ as the roots.
Hence, the required sum:
\begin{align*}
   \tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha &= -7
\end{align*}
From this equation, we can obtain several other identites, such as:
\begin{align*}
  \tan\alpha  +\tan2\alpha + \tan4\alpha  &= -\sqrt{7} \\
  \tan\alpha  \cdot \tan2\alpha \cdot \tan4\alpha  &= -\sqrt{7} \\
\end{align*}
And from $(6)$,
\begin{align*} \\
  \tan\left(\alpha\right)^2  +\tan\left(2\alpha\right)^2 + \tan\left(4\alpha\right)^2  &= 21 \\
   \left(\tan\alpha \tan2\alpha\right)^2 +\left(\tan2\alpha \tan4\alpha\right)^2 +\left(\tan4\alpha \tan\alpha\right)^2 &= 35 \\
   \tan\left(\alpha\right)^2  \cdot \tan\left(2\alpha\right)^2 \cdot \tan\left(4\alpha\right)^2  &= 7 \\
\end{align*}
and so on for powers of $4, 8$ etc.
