Hello, my first polynomial root problem!

$\begin{aligned} 36x^8-481x^6+1466x^4-481x^2+36 & = 0 & \small \color{#3D99F6} \text{Dividing both sides by }x^4 \\ 36x^4-481x^2+1466-\frac {481}{x^2}+\frac {36}{x^4} & = 0 \\ 36 x^4 + 72 + \frac {36}{x^4} - 481 x^2 - \frac {481}{x^2} + 1394 & = 0 \\ 36 {\color{#3D99F6} \left( x^2 + \frac 1{x^2} \right)}^2 - 481 {\color{#3D99F6}\left(x^2 + \frac 1{x^2} \right)} + 1394 & = 0 & \small \color{#3D99F6} \text{A quadratic equation of } x^2 + \frac 1{x^2} \\ \left(4\left(x^2 + \frac 1{x^2} \right)-17\right)\left(9\left(x^2 + \frac 1{x^2} \right)-82 \right) & = 0 \end{aligned}$

$\implies \begin{cases} 4x^2 -17 + \dfrac 4{x^2} = 0 & \implies 4x^4 -17x^2 + 4 = 0 \\ 9 x^2 - 82 + \dfrac 9{x^2} = 0 & \implies 9x^4 -82x^2 + 9 = 0 \end{cases}$

$\begin{aligned} 4x^4 -17x^2 + 4 & = 0 \\ (4x^2 - 1)(x^2-4) & = 0 \\ \implies x & = \pm \frac 12, \ \pm 2 \end{aligned}$

$\begin{aligned} 9x^4 -82x^2 + 9 & = 0 \\ (9x^2 - 1)(x^2-9) & = 0 \\ \implies x & = \pm \frac 13, \ \pm 3 \end{aligned}$

Therefore, $a=3$ , $b=-3$ and $673a + b + 1 = \boxed{2017}$