Evaluate this Trigonometric Expression: $\sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$ 
Evaluate
  $$ \sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$$  

I found the following    


*

*$\large{\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}=-\dfrac{1}{2}}$  

*$\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7} + \cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7} + \cos \frac{6\pi}{7}\times\cos \frac{2\pi}{7}}=-\dfrac{1}{2}$

*$\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7}=\dfrac{1}{8}}$   
Now, by Vieta's Formula's, $\large{\cos \frac{2\pi}{7}, \cos \frac{4\pi}{7} \: \text{&} \: \cos \frac{6\pi}{7}}$ are the roots of the cubic equation  
$$8x^3+4x^2-4x-1=0$$  
And, the problem reduces to finding the sum of cube roots of the solutions of this cubic.  
For that, I thought about transforming this equation to another one whose zeroes are the cube roots of the zeroes of this cubic by making the substitution   
$$x\mapsto x^3$$  
and getting another equation  
$$8x^9+4x^6-4x^3-1=0$$  
However, this new equation will have some extra roots too and we can't directly use Vieta's to get the desired sum.    
Also, it's given that the sum evaluates to a radical of the form  
$$\sqrt[3]{\frac{1}{d}(a-b\sqrt[b]{c})}$$  
where $a, b, c \: \text{&} \: d \in \mathbb Z$
Can somebody please help me with this question?
Thanks!

Try using this:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Let $a^3= \cos{\frac{2\pi}{7}},b^3= \cos{\frac{4\pi}{7}},c^3= \cos{\frac{6\pi}{7}} $.
