$(x+3y)^3=x^3+9x^2y+27(xy^2+y^3)=64 \to x+3y=4$ $y^3+(4-3y)y^2=2$ $2y^2-y^3=1$ $y^3-2y^2+1=0$ $y=1,\frac{1\pm \sqrt{5}}{2}$

Take $y=tx$ and the rest is easy calculation.

Too easy for germany 2015

KereMath wrote: Too easy for germany 2015 If that's your personal opinion, that's OK. But still, I would like to give two remarks: 1. The first problem in the final round usually is designed to be rather easy (= doable for a large portion of the participants). Compared with the corresponding first problems from 2016 and 2017, I wouldn't call this one too easy. 2. I remember at least one student who solved Problems 2 and 3 on that day but failed to solve Problem 1, so...

Tintarn wrote: KereMath wrote: Too easy for germany 2015 If that's your personal opinion, that's OK. But still, I would like to give two remarks: Yeah, when you aren't from Germany and decide what level the competition should be... From second equation $y\neq 0$. $$x=\frac{2-y^3}{y^2}$$$$10=\left(\frac{2-y^3}{y^2}\right)^3+9\left(\frac{2-y^3}{y^2}\right)^2\cdot y$$$$0=8(y^3-1)(y^6-4y^3-1)$$$y=1,x=1$ is a solution Now we consider $$y^6-4y^3-1=0$$which is $$y\in\left\lbrace\sqrt[3]{2+\sqrt{5}},\sqrt[3]{2-\sqrt{5}}\right\rbrace$$. We get solutions $$\left(\sqrt[3]{5(20-9\sqrt5)},\sqrt[3]{2+\sqrt{5}}\right),\left(\sqrt[3]{5(20+9\sqrt5)},\sqrt[3]{2-\sqrt{5}}\right)$$

Just as easy to solve for complex pairs $(x,y)$. $y^3+xy^2=2\implies x=\frac{2-y^3}{y^2}$ $x^3+9x^2y=10\implies\left(\frac{2-y^3}{y^2}\right)^3-9y\left(\frac{2-y^3}{y^2}\right)^2-10=0\implies (2-y^3)^3+9y^3(2-y^3)^2-10y^6=0$ $\implies y^9-5y^6+3y^3+1=0\implies(y^3-1)(y^6-4y^3-1)\implies(y^3-1)(y^3-\phi^3)(y^3+\phi^{-3})=0$ where $\phi$ is the golden ratio. $y\in\left\{1,\frac{-1\pm i\sqrt{3}}{2},\frac{1\pm\sqrt{5}}{2},\frac{-1-\sqrt{5}}{4}\pm\frac{\sqrt{15}+\sqrt{3}}{4}i,\frac{-1+\sqrt{5}}{4}\pm\frac{\sqrt{15}-\sqrt{3}}{4}i\right\}$ and respectively: $x\in\left\{1,\frac{-1\pm i\sqrt{3}}{2},\frac{5\mp 3\sqrt{5}}{2},\frac{-5+3\sqrt{5}}{4}\mp\frac{3\sqrt{15}-5\sqrt{3}}{4}i,\frac{-5-3\sqrt{5}}{4}\mp\frac{3\sqrt{15}+5\sqrt{3}}{4}i\right\}$

We got $$x^3+9x^2y=10$$$$27y^3+27xy^2=54$$then $$(x+3y)^3=64 \rightarrow (x+3y)=4$$so first equation we can transform to $$x^2(12-2x)=10 \rightarrow (x-1)(-2x^2+10x+10)=0$$$$x=1,\frac{5+3\sqrt5}{2},\frac{5-3\sqrt5}{2}$$then it is easy to find all solutions $$(x,y)=(1,1),(\frac{5-3\sqrt5}{2},\frac{1+\sqrt5}{2}),(\frac{5+3\sqrt5}{2},\frac{1-\sqrt5}{2})$$