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?


Using Dr. Sonnhard Graubner's answer, multiply everything by $27x^3$ and  you face an awful cubic equation in $x^3$ for which $\Delta=-1062882$ which means that there is only one real root.
Use the hyperbolic methid for that case and you will end with the "beautiful"
$$x^3=6 \left(a+b \cosh \left(\frac{1}{3}\cosh (c)\right)\right)$$ where
$$a=1+\sqrt[3]{2+\sqrt{2}}$$
$$b=\sqrt{3 \left(3+2 \sqrt[3]{2+\sqrt{2}}+\left(2+\sqrt{2}\right)^{2/3}\right)}$$
$$c=\frac{\sqrt{3} \left(8+\sqrt{2}+6 \sqrt[3]{2+\sqrt{2}}+4
   \left(2+\sqrt{2}\right)^{2/3}\right)}{2 \left(3+2
   \sqrt[3]{2+\sqrt{2}}+\left(2+\sqrt{2}\right)^{2/3}\right)^{3/2}}$$ Evaluated, this gives $$x^3=44.9381694189876 \implies x= ??? \implies y= ???$$
