Factoring $ x^4y^4 $ and $ x^3y^3 $ from the left-hand sides of the two equations, respectively, we get \[\begin{aligned} x^4y^4(x+y) &= 810, \\ x^3y^3(x^3+y^3) &= 945.\end{aligned}\]Let $ s = x+y $ and $ p = xy $. Then we can rewrite the given equations as \[\begin{aligned} p^4s &= 810, \\ p^3(s^3-3ps) &= 945,\end{aligned}\]using $ x^3+y^3 = (x+y)^3 - 3xy(x+y) = s^3 - 3ps $. Substituting $ s = 810/p^4 $ into the second equation, we get \[\begin{aligned} p^3\left(\frac{810^3}{p^{12}} - 3p \cdot \frac{810}{p^4}\right) &= 945 \\ \frac{810^3}{p^9} - 3 \cdot 810 &= 945 \\ p^9 &= \frac{810^3}{3 \cdot 810 + 945} = \frac{810^3}{15^3} = 54^3.\end{aligned}\]Thus $ p = \sqrt[3]{54}, $ and $ s = 810/p^4 = 810/(54\sqrt[3]{54}) = 15/\sqrt[3]{54} $. The quantity we want to compute is then \[\begin{aligned} 2x^3 + (xy)^3 + 2y^3 &= 2(s^3 - 3ps) + p^3 \\ &= 2s^3 - 6ps + p^3 \\ &= 2 \cdot \frac{15^3}{54} - 6 \cdot \sqrt[3]{54} \cdot \frac{15}{\sqrt[3]{54}} + 54 \\ &= 125 - 90 + 54 \\ &= \boxed{89}.\end{aligned}\]