Solve the equation $x^4+(x-1)(x^2-2x+2)=0$ With $x \in \mathbb{R}$. Solve the equation $$x^4+(x-1)(x^2-2x+2)=0$$

Idea $1$ : $(x^2-x+1)(x^2+2x-2)=0$
Idea $2$ :$(2x^2+x-1)^2=(3x-3)^2$
Idea $3$ :$<=>x^4(x-1)x^2-2(x-1)^2=0$
Let $y=x-1 \dots$
I need another way because I am collecting methods of solving with quartic equations
 A: Let $t=x-1+\frac1{x-1}$ to reduce the equation to a quadratic one
$$t^2 +5t+4=0$$
A: Let $t=x-1$. Then, $$x^4+(x-1)(x^2-2x+2)= x^4+(x-1)((x-1)^2+1)$$$$=(t+1)^4+t(t^2+1)=t^4+5t^3+6t^2+5t+1=0$$ Divide by $t^2$: $$\displaystyle \left(t^2+\frac{1}{t^2}\right)+5\left(t+\frac 1t\right)+6=0$$$$\implies \left(\left(t+\frac 1t\right)^2-2\right) +5\left(t+\frac 1t\right)+6=0$$ Let $u=\displaystyle t+\frac 1t$ so that $$u^2+5u+4=0.$$ This has solutions $u=-1,-4$ but will not admit the $-1$ solution because the range of $u=\displaystyle t+\frac 1t$ does not include $-1$. Thus, the roots are of $t^2-4t+1=0$. Add 1 to each root to get the values of x.
A: 
Solve $x^4+(x-1)(x^2-2x+2)=0$

You can solve this using Newton-Raphson :
$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}$$
Where \begin{align}&f(x)=x^4+x^3-3x^2+4x-2\\
&f'(x)=4x^3+3x^2-6x+4\\\end{align}
To guess $x_0$ :
\begin{align}&f(1)>0
\\&f(0)<0
\\\end{align}
So $x_0=1$ :
\begin{align}&x_1=0.5\\
&x_2=0.8\\
&x_3\approx0.6765\\
& \dots \\ 
&\dots \\ 
&\dots \\
&x_{22}\approx0.73205  \\\end{align}
The first Solution is $\boxed{x\approx0.73205}$ now Apply long division  :
$$\frac{x^4+x^3-3x^2+4x-2}{x-0.73205 }=x^3+1.73205 x^2-1.73205 x+2.73205$$
Now solve this equation with the same method you should get the $2$nd Solution if there is one
$$x^3+1.73205x^2-1.73205 x+2.73205 \approx 0$$
If there is a Solution apply long division again to find another one until you come to a function that doesn't have Solutions where $x \in\mathbb{R}$.
