It's not only algebra manipulation!

$\begin{aligned} x & = 1 + \frac 1{2+\frac 1{1+\frac 1{2+\ddots}}} = 1 + \frac 1{2+\frac 1x} = 1 + \frac x{2x+1} = \frac {3x+1}{2x+1} \end{aligned}$

$\begin{aligned} \implies x(2x+1) & = 3x+1 \\ 2x^2 + x & = 3x+1 \\ \color{#3D99F6} 2x^2 & \color{#3D99F6} = 2x + 1 \end{aligned}$

Now we need to find:

$\begin{aligned} \chi & = \left(4x^2-4\right)^2 + \left(8x^3-10\right)^4 & \small \color{#3D99F6} \text{Note that } 2x^2 = 2x + 1 \\ & = \left(2(2x+1)-4\right)^2 + \left(4x(2x+1)-10\right)^4 \\ & = \left(4x-2\right)^2 + \left(8x^2+4x-10 \right)^4 \\ & = 4\left(2x-1\right)^2 + \left(4(2x+1)+4x-10 \right)^4 \\ & = 4\left(4x^2-4x+1 \right) + \left(12x-6 \right)^4 \\ & = 4\left(2(2x+1)-4x+1 \right) + 1296 \left(2x-1 \right)^4 \\ & = 4\left(3 \right) + 1296 \left(3^2 \right) \\ & = \boxed{11676} \end{aligned}$

$\begin{aligned} x & = 1 + \dfrac{1}{2 + \dfrac{1}{1+ \dfrac{1}{2+ \cdots}}} \\ x & = 1 + \dfrac{1}{2+ \dfrac{1}{x}} \\ x-1 & = \dfrac{1}{2+ \dfrac{1}{x}} \\ x-1 & = \dfrac{x}{2x+1} \\ (2x+1)(x-1) & = x \\ 2x^2 -x -1 & = x \\ 2x^2 -2x -1 & =0 \end{aligned}$

By Al-Khawarizmi, we have $x = \dfrac{1+ \sqrt{3}}{2} \Rightarrow \text{Since} \space x>0$ .

We have $4x^2 = 4\left( \dfrac{(1+ \sqrt{3})^2}{4} \right)$

$4x^2 = 2\sqrt{3} + 4$ .

And, $8x^3 = 8\left( \dfrac{(1+\sqrt{3})^3}{8} \right)$

$8x^3 = 6\sqrt{3} + 10$ .

Hence, $\left( 4x^2 - 4 \right)^2 + \left( 8x^3 -10 \right)^4 = (2\sqrt{3})^2 + (6\sqrt{3} ) ^4 = 12 + 11664 = \boxed{11676}$