Let $$x$$ and $$y$$ be real numbers satisfying $$x^4y^5+y^4x^5=810$$ and $$x^3y^6+y^3x^6=945$$. Then the value of $$2x^3+x^3y^3+2y^3$$ is

Dividing the second equation by the first gives $$\frac{x^3+y^3}{xy(x+y)}=\frac{7}{6}$$. Since $$x^3+y^3=(x+y)(x^2-xy+y^2)$$, this gives $$6x^2+6y^2-13xy=0$$, or $$\left(3x-2y\right)\left(2x-3y\right)=0$$. Taking $$x=\frac{2y}{3}$$ in the first equation gives $$x^3=4$$, $$y^3=\frac{27}{2}$$ and $$x^3y^3=54$$, so the required value is $$2(4)+54+2\left(\frac{27}{2}\right)=89$$.