Solve the differential equation: (x3 - 3x2 + 2x + 1)/(x2 - 4x + 4) dx + (x2 + 4y - y2)/((4y + y2)^(1/3)) dy = 0

The expression provided is:

$$\left( \frac{x^3 - 3x^2 + 2x + 1}{x^2 - 4x + 4} \right) dx \left( \frac{dx}{dy} \right) + \left( \frac{x^2 + 4y - y^2}{\sqrt[3]{4y + y^2}} \right)$$

The denominator of the first term is $x^2 - 4x + 4$, which can be factored as $(x-2)^2$.
We perform polynomial long division for $\frac{x^3 - 3x^2 + 2x + 1}{x^2 - 4x + 4}$:

$$\begin{array}{c|cc cc} \multicolumn{2}{r}{x} & +1 \\ \cline{2-5} x^2-4x+4 & x^3 & -3x^2 & +2x & +1 \\ \multicolumn{2}{r}{-(x^3} & -4x^2 & +4x) \\ \cline{2-4} \multicolumn{2}{r}{0} & x^2 & -2x & +1 \\ \multicolumn{2}{r}{} & -(x^2 & -4x & +4) \\ \cline{3-5} \multicolumn{2}{r}{} & 0 & 2x & -3 \\ \end{array}$$

So, the rational function simplifies to $x + 1 + \frac{2x - 3}{x^2 - 4x + 4}$, or $x + 1 + \frac{2x - 3}{(x-2)^2}$.

Substituting the simplified form into the original expression gives:

$$\left( x + 1 + \frac{2x - 3}{(x-2)^2} \right) dx \left( \frac{dx}{dy} \right) + \left( \frac{x^2 + 4y - y^2}{\sqrt[3]{4y + y^2}} \right)$$

The simplified expression is:

$$\boxed{\left( x + 1 + \frac{2x - 3}{(x-2)^2} \right) dx \left( \frac{dx}{dy} \right) + \left( \frac{x^2 + 4y - y^2}{\sqrt[3]{4y + y^2}} \right)}$$