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!
 A: 
let  $$x=\sqrt[3]{\cos{\dfrac{2\pi}{7}}},y=\sqrt[3]{\cos{\dfrac{4\pi}{7}}},z=\sqrt[3]{\cos{\dfrac{6\pi}{7}}},$$
  then we have
  $$\begin{cases}
x^3+y^3+z^3=-\dfrac{1}{2}\\
(xy)^3+(yz)^3+(xz)^3=-\dfrac{1}{2}\\
(xyz)^3=\dfrac{1}{8}
\end{cases}$$
  use this identity
  $$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc$$
  so
  $$\begin{cases}
(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz=-\dfrac{1}{2}\\
(xy+yz+xz)^3-3(xy+yz+xz)[xyz(x+y+z)]+3x^2y^2z^2=-\dfrac{1}{2}\\
xyz=\dfrac{1}{2}
\end{cases}$$
  let $$u=x+y+z, v=xy+yz+xz$$
  then we have
  $$\begin{cases}
u^3-3uv+2=0\\
4v^3-6uv+5=0
\end{cases}$$
  so we have
  $$\Longrightarrow 4v^3-2u^3+1=0, v=\dfrac{u^3+2}{3u}$$
  so
  $$4\left(\dfrac{u^3+2}{3u}\right)^3-2u^3+1=0\Longrightarrow 4u^9-30u^6+75u^3+32=0$$
  let $t=u^3$,so we have
  $$4t^3-30t^2+75t+32=0$$
  let $t=\dfrac{5}{2}-a$,then
  $$4\left(\dfrac{5}{2}-a\right)^3-30\left(\dfrac{5}{2}-a\right)^2+75\left(\dfrac{5}{2}-a\right)+32=0$$
  $$\Longrightarrow 4a^3=\dfrac{189}{2}\Longrightarrow a=\dfrac{3\sqrt[3]{7}}{2}$$
  so
  $$u=x+y+z=\sqrt[3]{\cos{\dfrac{2\pi}{7}}}+\sqrt[3]{\cos{\dfrac{4\pi}{7}}}+\sqrt[3]{\cos{\dfrac{6\pi}{7}}}=\sqrt[3]{\dfrac{1}{2}\left(5-3\sqrt[3]{7}\right)}$$

A: 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
A: 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}} $.
