Asymmetric Polynomial

This is not a complete solution. First I made the substitution $x=2-\dfrac{1}{t+2}$ to obtain $\left(2-\dfrac{1}{t+2}\right)^3-7\left(2-\dfrac{1}{t+2}\right)+7=0$ . After expanding and multiply both sides by $(t+2)^3$ we obtain $t^3+t^2-2t-1=0$ . Now let $t=y+\dfrac{1}{y}$ :

$\left(y+\dfrac{1}{y}\right)^3+\left(y+\dfrac{1}{y}\right)^2-2\left(y+\dfrac{1}{y}\right)-1=0$

Do the same, expand and multiply both sides by $y^3$ :

$y^6+y^5+y^4+y^3+y^2+y+1=0$

$y^7=1$ , $y \neq 1$

So $y=e^\frac{2 \pi k i}{7}$ for $k \in [1,6]$ .

Now, $t=e^\frac{2 \pi k i}{7}+e^{-\frac{2 \pi k i}{7}}=2\cos \left(\dfrac{2 \pi k}{7}\right)$ , this time for $k \in [1,3]$ because for $k \in [4,6]$ we obtain the same.

And finally, $x=2-\dfrac{1}{2+2\cos \left(\dfrac{2 \pi k}{7}\right)}=2-\dfrac{1}{4} \sec^2 \left(\dfrac{\pi k}{7}\right)$ .

Obtain $a$ , $b$ and $c$ with the given condition:

$a=2-\dfrac{1}{4} \sec^2\left(\dfrac{3\pi}{7}\right) \\ b=2-\dfrac{1}{4} \sec^2\left(\dfrac{2\pi}{7}\right) \\ c=2-\dfrac{1}{4} \sec^2\left(\dfrac{\pi}{7}\right)$

But how can we find the asked value from here? I tried to substitute them but I got stuck.

Try using newton's sums

I must confess I solved it, by using the peasant way , calculating the roots of a depleted cubic.

I am sure the bright brains, plentiful around in this website, will use a clever and skillful manipulation of Cardano-Vietta relations to come out with a more elegant procedure so far elusive to me.