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!

My solution:
Let $\alpha= \sqrt [3] {\cos \frac{2\pi}{7}}, \beta=\sqrt [3] {\cos \frac{4\pi}{7}}, \gamma=\sqrt [3] {\cos \frac{8\pi}{7}}$ so $(x-{\alpha}^3)(x-{\beta}^3)(x-{\gamma}^3)$ $=8x^3+4x^2-4x-1$.
hence we get $${\alpha}^3+{\beta}^3+{\gamma}^3=-\frac{1}{2} \tag{1}$$ $${\alpha}^3{\beta}^3+ {\beta}^3{\gamma}^3+ {\gamma}^3{\alpha}^3=\frac{-1}{2} \tag{2}$$
$${\alpha}^3{\beta}^3{\gamma}^3=\frac{1}{8} \Longrightarrow {\alpha}{\beta}{\gamma}=\frac{1}{2}. \tag{3}$$
Let $\mathcal{P}={\alpha}+{\beta}+{\gamma}$ and $\mathcal{Q}={\alpha}{\beta}+ {\beta}{\gamma}+ {\gamma}{\alpha}$.
From $(2)$, ${\alpha}^3+{\beta}^3+{\gamma}^3-3{\alpha}{\beta}{\gamma}=\mathcal{P}(\mathcal{P} ^2-3\mathcal{Q} ) \Longrightarrow \mathcal{P} ^3+2=3 \mathcal{P} \mathcal{Q} \ \tag{4}.$
From $(1), (3)$ and $\sum ( {\beta} ^2 {\gamma}+{\beta} {\gamma} ^2) =\mathcal{P} (\mathcal{P} ^2- 2 \mathcal{Q} )- \sum {\alpha}^3$ we get $\mathcal{Q} ^3=\sum {\beta}^3 {\gamma} ^3+6 {\alpha}^2 {\beta}^2 {\gamma}^2 +\frac{3}{2} ( \mathcal{P} (\mathcal{P} ^2- 2  \mathcal{Q} )+\frac{1}{2}),$ so combine with $(4)$ we get $$\mathcal{Q} ^3=\frac{7}{4}+\frac{3}{2} \mathcal{P} ^3-3 \mathcal{P} \mathcal{Q} =\frac{-1}{4}+\frac{1}{2} \mathcal{P} ^3 \Longrightarrow \mathcal{Q} ^3=\frac{1}{4} ( 2 \mathcal{P} ^3 -1). \tag{5}$$
From $(4), (5)$, $(\mathcal{P} ^3 +2)^3=27 \mathcal{P} ^3 \mathcal{Q} ^3=\frac{27}{4} \mathcal{P} ^3 ( 2 \mathcal{P} ^3 -1 ) \Longrightarrow 8 (\mathcal{P} ^3 )^3-60 (\mathcal{P} ^3)^2+150 \mathcal{P} ^3+64=0  ,$ hence we get:
$$(2\mathcal{P} ^3 -5)^3 = -189 \Longrightarrow \mathcal{P} ^3=\frac{5-3\sqrt [3]{7}}{2} \Longrightarrow \sqrt [3] {\cos \frac{2\pi}{7}}+\sqrt [3] {\cos \frac{4\pi}{7}}+\sqrt [3] {\cos \frac{8\pi}{7}}=\mathcal{P}$$
$$= \sqrt [3]{\frac{5-3\sqrt [3]{7}}{2}}.$$
Q.E.D
