<s>[INST] Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$ Solve the system of equations for   $x,y \in  \mathbb{R}$

$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}  $
$ y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}$

I just rearranged between those equations and get $ \frac{y^{3}-9}{x^{3} -6} = (\sqrt{2}+2)^{\frac{1}{3}}$ then I don't know how to deal with it.
Please give me a hint or relevant theorem to solve the equation.
Thank you, and I appreciate any help.

Furthermore I get an idea
how about we subtract two equation and get
$y^{3}-x^{3} = 3 +  3xy((\sqrt{2}+2)^{\frac{1}{3}} -1)  - (3(\sqrt{2}+2)^{\frac{2}{3}} - 3(\sqrt{2}+2)^{\frac{1}{3}})$
$(y-x)(x^2+xy+y^2)= 3[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)]$
$y-x = 3$ and $x^2 +xy+y^2 =[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $
or,
$y-x = [1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $ and $x^2 +xy+y^2 = 3$
Am I on the right track?

 [/INST] $$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}$$
$$y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}} $$
Let $A=\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}$, then re-write the equations as
$$x^{3} = 6+ 3xy - 3A\tag{1}$$
$$y^{3} = 9 + 3xyA - 3A^2\tag{2} $$
Solving (1) for $3xy$ 
$$ 3xy=x^3+3A-6 \tag{3}$$
Substituting (3) into (2) gives
$$ y^3=Ax^3-6A+9 \tag{4}$$
Cubing (3) and combining with (4) gives
$$ 27x^3(Ax^3-6A+9)=(x^3+3A-6)^3 $$
Let $x^3=u,\,6A-9=B$, and $3A-6=C$. Then equation (5) becomes
$$ 27u(Au-B)=(u+C)^3 $$
$$ u^3+(3C-27A)u^2+(3C^2+27B)u+C^3=0 $$
Solving for $c$ and the cube roots of $c$ there is only one real solution $(3.555,4.074)$ [for which I cheated and used desmos].

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