Let $\alpha = \left(1 - 2\cos\left(\frac{\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{3\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{9\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{27\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{81\pi}{11}\right)\right)$. Then the value of $5 - \alpha^{2}$ is _______.

$\alpha = \left(1 - 2\cos\left(\frac{\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{3\pi}{11}\right)\right)$ $\qquad \left(1 - 2\cos\left(\frac{9\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{27\pi}{11}\right)\right)\left(1 - 2\cos\left(\frac{81\pi}{11}\right)\right)$ $\text{To find: } 5 - \alpha^2$ $1.\ \text{set } x_k = 2\cos\frac{3^k\pi}{11}$ $2.\ \text{show each } x_k \text{ is a root of } f_{11}(x)+2 = 0$ $3.\ \text{factor and find } \alpha^2$ $x_k = 2\cos\frac{3^k\pi}{11}, \quad k = 0,1,2,3,4$ $\alpha = \prod_{k=0}^{4} (1 - x_k)$ $f_n(2\cos\theta) = 2\cos(n\theta)$ $f_0 = 2,\quad f_1 = x,\quad f_{n+1} = x f_n - f_{n-1}$ $f_{11}(x_k) = 2\cos\!\left(11\cdot\frac{3^k\pi}{11}\right) = 2\cos(3^k\pi)$ $3^k \text{ is odd} \ \Rightarrow\ 2\cos(3^k\pi) = -2$ $f_{11}(x)+2 = 0 \text{ has these five } x_k \text{ as roots}$ $x_k \in \left\{2\cos\frac{k\pi}{11} : k = 1,3,5,7,9\right\}$ $f_{11}(x)+2 = (x+2)\,P(x)^2$ $P(x) = \prod_{k\in\{1,3,5,7,9\}} \left(x - 2\cos\frac{k\pi}{11}\right)$ $\alpha = \prod_{k=0}^{4}(1 - x_k) = P(1)$ $\alpha^2 = P(1)^2$ $f_{11}(1)+2 = 3\,P(1)^2$ $\Rightarrow\ P(1)^2 = \frac{f_{11}(1)+2}{3}$ $1 = 2\cos\frac{\pi}{3} \ \Rightarrow\ f_{11}(1) = 2\cos\frac{11\pi}{3} = 2\cos\frac{5\pi}{3} = 1$ $\alpha^2 = P(1)^2 = \frac{1+2}{3} = 1$ $5 - \alpha^2 = 5 - 1 = 4$ $\boxed{4}$