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

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.
