jmc

Factoring $x^4{y}^4$ and $x^3{y}^3$ from the left-hand sides of the two equations, respectively, we get $$\begin{array}{rl}{x}^4{y}^4(x+y)& =810,\\ x^3{y}^3(x^3+y^3)& =945.\end{array}$$Let $s=x+y$ and $p=xy.$ Then we can rewrite the given equations as $$\begin{array}{rl}{p}^{4}s& =810,\\ p^3(s^3-3ps)& =945,\end{array}$$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 $$\begin{array}{rl}{p}^3\left(\frac{810^3}{p^{12}}-3p\cdot \frac{810}{p^4}\right)& =945\\ \frac{810^3}{p^9}-3\cdot 810& =945\\ p^9& =\frac{810^3}{3\cdot 810+945}=\frac{810^3}{15^3}=54^3.\end{array}$$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 $$\begin{array}{rl}2x^3+(xy{)}^3+2y^3& =2(s^3-3ps)+p^3\\ & =2s^3-6ps+p^3\\ & =2\cdot \frac{15^3}{54}-6\cdot \sqrt[3]{54}\cdot \frac{15}{\sqrt[3]{54}}+54\\ & =125-90+54\\ & =\boxed{89}.\end{array}$$