<s>[INST] Find roots of $x^4 -6x^3 + 12x^2 - 12 x + 4 = 0$ the original equation is:
$$(x^2 + 2)^2 -6x(x^2+2) + 8x^2=0.$$
cannot see how to go solving this. I tried following way to factorise:
$$(x^2+2)(x^2-6x+2) + 8x^2 = 0.$$
But this has no help to solve. 
Thank you people, but I need the thinking process, not the answer.
 [/INST] 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 $x\in\{2+\sqrt2,2-\sqrt2\}$.
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