Root finder

Let $a=(x^2-2x+5)^2-8(x^2-2x+5)+25$ , then $a^2-18a+81=0,a=9$

Let $b=x^2-2x+5$ , then $a=b^2-8b+25=9,b=4$

$b=x^2-2x+5=4,x=1$

$\begin{aligned} {\color{#D61F06}\left({\color{#3D99F6}(x^2-2x+5)^2 - 8(x^2-2x+5)+25}\right)^2} - 18{\color{#D61F06}\left({\color{#3D99F6}(x^2-2x+5)^2 - 8(x^2-2x+5)+25}\right)} + 81 & = 0 & \small \color{#3D99F6} \text{Let } u = x^2 - 2x + 5 \\ {\color{#D61F06}\left({\color{#3D99F6}u^2 - 8u+25}\right)^2} - 18{\color{#D61F06}\left({\color{#3D99F6}u^2 - 8u+25}\right)} + 81 & = 0 & \small \color{#3D99F6} \text{Let } t = u^2 - 8u + 25 \\ {\color{#D61F06}t^2} - 18{\color{#D61F06}t} + 81 & = 0 \\ \implies (t-9)^2 & = 0 & \small \color{#3D99F6} \implies t =9 \\ \implies u^2 - 8u + 25 & = 9 \\ u^2 - 8u + 16 & = 0 \\ (u-4)^2 & = 0 & \small \color{#3D99F6} \implies u = 4 \\ \implies x^2-2x+5 & = 4 \\ x^2 - 2x + 1 & = 0 \\ (x-1)^2 & = 0 \\ \implies x & = \boxed 1 \end{aligned}$

$\begin{aligned} \{(x^2 - 2x+ 5)^2 - 8(x^2 - 2x + 5) + 25\}^2 - 18\{(x^2-2x+5)^2 - 8(x^2 - 2x + 5) + 25\} + 81 & = 0 \\ \Rightarrow (y^2 - 8y + 25)^2 - 18(y^2 - 8y +25) + 81 & = 0 ~~~~ [x^2 - 2x + 5 = y] \\ \Rightarrow z^2 - 18z + 81 & = 0 ~~~~ [ y^2 - 8y + 25 = z] \\ \Rightarrow z^2 - 9z - 9 z + 81 & = 0 \\ \Rightarrow z(z-9) - 9(z-9) & = 0 \\ \Rightarrow (z - 9)(z - 9) & = 0 \\ \implies z = 9 \\ \end{aligned}$

Now,

$\begin{aligned} z & = 9 \\ \Rightarrow y^2 - 8y + 25 & = 9 \\ \Rightarrow y^2 - 8y + 16 & = 0 \\ \Rightarrow y^2 - 4y - 4y + 16 & =0 \\ \Rightarrow y(y - 4) - 4(y - 4) & = 0 \\ \Rightarrow (y - 4)(y - 4) & = 0 \\ \implies y = 4. \\ \end{aligned}$

Hence,

$\begin{aligned} y & = 4 \\ \Rightarrow x^2 - 2x + 5 &= 4 \\ \Rightarrow x^2 - 2x + 1 & = 0 \\ \Rightarrow x^2 - x - x + 1 & = 0 \\ \Rightarrow x(x -1) - (x - 1) & = 0 \\ \Rightarrow (x -1)(x -1) & = 0 \\ \implies x = \boxed 1 \\ \end{aligned}$