imc

$$x+\frac{x^3}{y^2}+\frac{y^3}{x^2}+y=x+\frac{x^3}{y^2}+y+\frac{y^3}{x^2}=\frac{x^3}{x^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}+\frac{x^3}{y^2}$$ Continuing to combine $$\frac{x^3+y^3}{x^2}+\frac{x^3+y^3}{y^2}=\frac{(x^2+y^2)(x^3+y^3)}{x^2{y}^2}=\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2{y}^2}$$ From the givens, it can be concluded that $x^2{y}^2=4$. Also, $$(x+y{)}^2=x^2+2xy+y^2=16$$ This means that $x^2+y^2=20$. Substituting this information into $\frac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2{y}^2}$, we have $\frac{(20)(4)(22)}{4}=20\cdot 22=\boxed{440}$.