Inspired by Paul Patawaran

$\begin{aligned} 4b^2 - 16b + 4 & = 0 & \small \color{#3D99F6} \text{Dividing both sides by }4 \\ b^2 - 4b + 1 & = 0 & \small \color{#3D99F6} \text{Dividing both sides by }b \\ b - 4 + \frac 1b & = 0 & \small \color{#3D99F6} \text{Rearranging} \\ \implies \color{#3D99F6} b + \frac 1b & \color{#3D99F6} = 4 \\ \left(b + \frac 1b\right)^2 & = 4^2 \\ b^2 + 2 + \frac 1{b^2} & = 16 \\ \implies \color{#3D99F6}b^2 + \frac 1{b^2} & \color{#3D99F6} = 14 \\ \left(b + \frac 1b\right) \left(b^2 + \frac 1{b^2} \right) & = 4 \times 14 \\ b^3 + b + \frac 1b + \frac 1{b^3} & = 56 \\ b^3 + 4 + \frac 1{b^3} & = 56 \\ \implies \color{#3D99F6}b^3 + \frac 1{b^3} & \color{#3D99F6} = 52 \\ \left(b^2 + \frac 1{b^2} \right) \left(b^3 + \frac 1{b^3} \right) & = 14 \times 52 \\ b^5 + b + \frac 1b + \frac 1{b^5} & = 728 \\ b^5 + 4 + \frac 1{b^5} & = 728 \\ \implies \color{#3D99F6}b^5 + \frac 1{b^5} & \color{#3D99F6} = 724 \\ \left(b^3 + \frac 1{b^3} \right)^2 & = 52^2 \\ b^6 + 2 + \frac 1{b^6} & = 2704 \\ \implies \color{#3D99F6}b^6 + \frac 1{b^6} & \color{#3D99F6} = 2702 \end{aligned}$

Now, we have:

$\begin{aligned} \left(b^6 + b^5 \right)\left(\frac 1{b^{11}} + 1 \right) & = \frac 1{b^5} + b^6 + \frac 1{b^6} + b^5 \\ & = b^5 + \frac 1{b^5} + b^6 + \frac 1{b^6} \\ & = 724 + 2702 \\ & = \boxed{3426} \end{aligned}$