Can you guess it?

Make the substitution $x=\dfrac{t}{t-2}$ :

$7\left(\dfrac{t}{t-2}\right)^3+7\left(\dfrac{t}{t-2}\right)^2-7\left(\dfrac{t}{t-2}\right)+1=0$

Expand everything and multiply both sides by $\dfrac{(t-2)^3}{8}$ :

$t^3+t^2-2t-1=0$

That is the minimal polynomial of $2\cos \left(\dfrac{2 \pi k}{7}\right)$ , for $k\in[1,3]$ . To prove that just let $z^7=1$ , that implies that $z=\cos\left(\dfrac{2 \pi k}{7}\right)+i\sin\left(\dfrac{2 \pi k}{7}\right)$ and $(z-1)(z^6+z^5+z^4+z^3+z^2+z+1)=0$ . Now let $t=z+\dfrac{1}{z}=2 \cos\left(\dfrac{2 \pi k}{7}\right)$ . Dividing both sides of the equation in $z$ by $z^3$ we obtain; $z^3+z^2+z+1+\dfrac{1}{z}+\dfrac{1}{z^2}+\dfrac{1}{z^3}=0$ . After a little algebraic manipulation we obtain $\left(z+\dfrac{1}{z}\right)^3+\left(z+\dfrac{1}{z}\right)^2-2\left(z+\dfrac{1}{z}\right)-1=0$ , so $t^3+t^2-2t-1=0$ .

Then, our roots are:

$x=\dfrac{t}{t-2}=\dfrac{1}{1-\frac{2}{t}}=\dfrac{1}{1-\sec\left(\dfrac{2 \pi k}{7}\right)}$ .

By the given conditions, after adjusting the angles in the range $(0, \frac{\pi}{2})$ , we conclude that:

$x_1=\dfrac{1}{1-\sec\left(\dfrac{2 \pi}{7}\right)} \\ x_2=\dfrac{1}{1+\sec\left(\dfrac{3 \pi}{7}\right)} \\ x_3=\dfrac{1}{1+\sec\left(\dfrac{\pi}{7}\right)}$

Hence, $a=1$ , $b=7$ , $m=2$ , $n=3$ , $p=1$ and $2b-(a+m+n+p)=\boxed{7}$ .

As the problem's title suggests, I actually guess the solution. The below is how I did it with a spreadsheet.

Let $f(x) = 7x^3+7x^2-7x+1$ , Then $x_1$ , $x_2$ and $x_3$ can be found by Newton's method $x_{n+1} = x_n - \dfrac {f(x_n)}{f'(x_n)} = x_n - \dfrac {7x_n^3+7x_n^2-7x_n+1}{21x_n^2+14x_n-7}$

The three roots are found to be $x_1 = -1.655970555$ , $x_2=0.182018097$ and $x_3 = 0.473952458$ .

Let $\theta_1 = \dfrac {m\pi}{b}$ , $\theta_2 = \dfrac {n\pi}{b}$ and $\theta_3 = \dfrac {p\pi}{b}$ ; then, we have:

$x_1 = \dfrac {a}{a - \sec {\theta_1}} = \dfrac {a}{a - \dfrac {1}{\cos {\theta_1}}} = \dfrac {a\cos{\theta_1}}{a\cos{\theta_1}-1} \quad \Rightarrow a\cos{\theta_1} = \dfrac {x_1}{x_1-1}$

Similarly $a\cos{\theta_2} = \dfrac {x_2}{1-x_2}$ and $a\cos{\theta_3} = \dfrac {x_3}{1-x_3}$

Assuming $a = 1$ , we find that:

\(\begin{array} {} \cos{\theta_1} = 0.623489802 & \theta_1 = 0.897597901 & \dfrac {\theta_1}{\pi} = \dfrac {m}{b} = 0.285714286 \\ \cos{\theta_2} = 0.222520934 & \theta_2 = 1.346396852 & \dfrac {\theta_2}{\pi} = \dfrac {n}{b} = 0.428571429\\ \cos{\theta_3} = 0.900968868 & \theta_3 = 0.448798951 & \dfrac {\theta_3}{\pi} = \dfrac {p}{b} = 0.142857143 \end{array} \)

By try and error, we find that:

\(\begin{array} {} \dfrac {\theta_1}{\pi} \times 7 = 0.285714286 \times 7 = 2 & \Rightarrow b = 7 \quad m = 2 \\ \dfrac {\theta_2}{\pi} \times 7 = 0.428571429 \times 7 = 3 &\Rightarrow b = 7 \quad n = 3 \\ \dfrac {\theta_3} {\pi} \times 7 = 0.142857143 \times 7 = 1 & \Rightarrow b = 7 \quad p = 1 \end{array} \)

Therefore, $2b - (a+m+n+p) = 2\times 7 - (1+2+3+1) = \boxed{7}$