AIME 2015 Problem 14

$\begin{cases} x^4y^5+y^4x^5 = (xy)^4(x+y) = 810 &...(1) \\ x^3y^6+y^6x^3 = (xy)^3 (x^3+y^3) =(xy)^3(x+y) (x^2+y^2-xy) = 945 &...(2) \end{cases}$

$\begin{aligned} \frac{(2)}{(1)}: \quad \frac{x^2+y^2-xy}{xy} & = \frac{945}{810} \\ x^2+y^2-xy & = \frac{7}{6} xy \\ x^2+y^2 & = \frac{7}{6} xy + xy = \frac{13}{6}xy \\ \text{Since } (x+y)^2 & = x^2 + y^2 + 2xy = \frac{13}{6}xy + 2xy = \frac{25}{6}xy \\ \Rightarrow x + y & = \frac{5}{\sqrt{6}}\sqrt{xy} \\ (1): \quad \quad (xy)^4(x+y) & = 810 \\ (xy)^4 \left( \frac{5}{\sqrt{6}}\sqrt{xy} \right) & = 810 \\ (xy)^\frac{9}{2} & = 162\sqrt{6} \\ \Rightarrow (xy)^3 & = \left(162\sqrt{6}\right)^{\frac{2}{9}\times 3} = \sqrt[3]{2^33^9} = 54 \\ (2): \quad \quad (xy)^3 (x^3+y^3) & = 945 \\ 54 (x^3+y^3) & = 945 \\ \Rightarrow x^3 + y^3 & = \frac{35}{2} \\ \Rightarrow 2x^3 + (xy)^3 + 2y^3 & = 35 + 54 = \boxed{89} \end{aligned}$

I wonder if my solution is any better...

Assume without loss of happiness and generality that $y \geq x$

Then, from the above problem, $x^4y^5+x^5y^4 = 810$ , $(x^4y^4)(x+y) = 810$ , $x+y = \frac{810}{x^4y^4}$

Similarly, $x^3y^6+x^6y^3=945$ , $(x^3y^3)(x+y)(x^2-xy+y^2)=945$ , and since $x+y = \frac{810}{x^4y^4}$ ,

$(x^3y^3)(x+y)(x^2-xy+y^2)=945$

$\frac{810 \cdot (x^3y^3)(x^2-xy+y^2)}{x^4y^4}=945$

$\frac{x^2-xy+y^2}{xy} = \frac {945}{810}$

Which gives us $6x^2-13xy+6y^2=0$

$(3x-2y)(2x-3y) = 0$

Since $y \geq x$ , $3x-2y = 0$ , and $y = \frac{3x}{2}$

Substituting, $(x^4y^4)(x+y) = 810$ , $(\frac{81x^8}{32})(\frac{5x}{2} = 810)$

This gives us $x^9 = 64, x^3 = 4$ , and similarly, $y^3 = \frac{27 x^3}{8}, y^3 = \frac{27}{2}$

Plugging in the values, $2x^3+x^3y^3+2y^3 = 2(4)+(4)(\frac{27}{2})+2(\frac{27}{2})=89$

As conclusion, the answer is 89, and since my solution is longer than Chew-Seong Cheong's, my solution is no better T.T </3

$x^4*y^4(x+y)=810=3^3*5*6..........(1)\\ x^3*y^3*(x^3+y^3) =x^3*y^3*(x+y)(x^2-xy+y^2)=3^3*5*7..........(2)\\ \therefore\ \dfrac{(2)}{(1)}= \dfrac{x^2-xy+y^2}{xy}=\frac 7 6.\\ \implies\ \ \ x^2-\frac {13} 6*xy+y^2.\\ WLOG\ y\geq x,\ solving\ the\ quadratic\ \ y=\frac 3 2*x.\\ Substituting\ this\ in\ (2),\ \ \ x^9*(\frac 3 2)^3*(1+ (\frac 3 2)^3\ )=3^3*5*7\\ \therefore\ x^9=8*8,\ \ \implies\ \color{#3D99F6}{x^3=4,\ \ and\ \ y^3}=4*(\frac 3 2)^3*x^3=\color{#3D99F6}{\frac{27} 2*x^3.}\\ \therefor\ \ 2x^3+xy+2y^3=2*(4+\frac {27} 2)+4*\frac {27} 2= \Large\ \ \ \color{#D61F06}{89}.$