Find roots of $x^4 -6x^3 + 12x^2 - 12 x + 4 = 0$

In the original version, let $y=x^2+2$. Then, we have $y^2-6xy+8x^2=0$, which gives us $(y-2x)(y-4x)=0$.

It's $$\left(\frac{x^2+2}{x}\right)^2-6\left(\frac{x^2+2}{x}\right)+8=0,$$ which gives $$\frac{x^2+2}{x}=2$$ and $x\in\{1+i,1-i\}$ or $$\frac{x^2+2}{x}=4,$$ which gives $x\in\{2+\sqrt2,2-\sqrt2\}$.

your equation is equivalent to $$(x^2-4x+2)(x^2-2x+2)=0$$