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Real numbers $x$ and $y$ satisfy $x+y=4$ and $x\cdot y=-2$. What is the value of $x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y$? (2020 AMC 10A Problem, Question \#14) 
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$x+\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}+y=x+\frac{y^{3}}{x^{2}}+y+ \frac{x^{3}}{y^{2}}=\frac{x^{3}}{x^{2}}+\frac{y^{3}}{x^{2}}+\frac{y^{3}}{y^{2}}+\frac{x^{3}}{y^{2}}$  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 $\frac{20\times 4\times22}{4}=440$. 