$VT=4x^3+4$
$VP=x^3+(x^2-2x+2)\sqrt{x^2-2x+2}+3.(x^2-2x+2).x+3.\sqrt{x^2-2x+2}.x^2$$=4x^3-6x^2+6x+\sqrt{x^2-2x+2}(4x^3-2x+2)$
Nên $pt\Leftrightarrow 3x^2-3x+2=\sqrt{x^2-2x+2}(2x^2-x+1)$
$\Leftrightarrow (3x^2-3x+2)^2=(x^2-2x+2)(2x^2-x+1)^2$
$\Leftrightarrow (x-1)^3(2x^3+1)=0\Leftrightarrow \boxed{x=1 \quad \text{hoặc} \quad x=\frac{-1}{\sqrt[3]2}}$