Not really complex

$\begin{aligned} 4x^2 + 2x + 1 & = 0 & \small \color{#3D99F6} \text{Multiplying both sides by }2x \\ 8x^3 + 4x^2 + 2x & = 0 & \small \color{#3D99F6} \text{Adding 1 to both sides} \\ 8x^3 + {\color{#3D99F6}4x^2 + 2x + 1} & = 1 \\ 8x^3 + {\color{#3D99F6}0} & = 1 \\ \implies x^3 & = \frac 18 \end{aligned}$

Therefore, we have:

$\begin{aligned} S & = x^3+x^6+x^9+x^{12}+\cdots \\ & = \frac 18 + \left(\frac 18 \right)^2 + \left(\frac 18 \right)^3 + \left(\frac 18 \right)^4 + \cdots \\ & = \frac 18 \left(\frac 1{1-\frac 18}\right) \\ & = \frac 17 \end{aligned}$

$\implies a + b = 1 + 7 = \boxed{8}$

$4x^2+2x+1=0\quad \Leftrightarrow (2x-1)(4x^2+2x+1)=0$

This means that $8x^3-1=0$ , or $x^3=\frac{1}{8}$

The sum $x^3+x^6+x^9+\ldots$ is a geometric progression wrt $x^3$ and with common ratio $\frac{1}{8}$ , so $x^3+x^6+x^9+\ldots=\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac{1}{7}$ , so $a+b=1+7=8$