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Real numbers $x$ and $y$ satisfy $x+y=4$ and $x \cdot y=-2$. What is the value of $$ x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y? $$ 
Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Proportional Equations
$$ x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y=x+\frac{x^{3}}{y^{2}}+y+\frac{y^{3}}{x^{2}}=\frac{x^{3}}{x^{2}}+\frac{y^{3}}{x^{2}}+\frac{y^{3}}{y^{2}}+\frac{x^{3}}{y^{2}} $$ Continuing to combine $$ \frac{x^{3}+y^{3}}{x^{2}}+\frac{x^{3}+y^{3}}{y^{2}}=\frac{\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)}{x^{2} y^{2}}=\frac{\left(x^{2}+y^{2}\right)(x+y)\left(x^{2}-x y+y^{2}\right)}{x^{2} y^{2}} $$ From the givens, it can be concluded that $x^{2} y^{2}=4$. Also, $$ (x+y)^{2}=x^{2}+2 x y+y^{2}=16 $$ This means that $x^{2}+y^{2}=20$. Substituting this information into $\frac{\left(x^{2}+y^{2}\right)(x+y)\left(x^{2}-x y+y^{2}\right)}{x^{2} y^{2}}$, we have $\frac{(20)(4)(22)}{4}=20 \cdot 22=$ 440. 