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?

 A: 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 $$x^3=44.9381694189876 \implies x= ??? \implies y= ???$$
A: Solving your first equation for $y$ we get
$$y=\frac{x^3+3 \sqrt[3]{2+\sqrt{2}}-6}{3 x}$$
Plugging this in the second equation we get
$$\frac{x^6}{27}-\frac{2}{3}
   \left(1+\sqrt[3]{2+\sqrt{2}}\right)
   x^3+\frac{-6+\sqrt{2}+12 \sqrt[3]{2+\sqrt{2}}-6
   \left(2+\sqrt{2}\right)^{2/3}}{x^3}-2
   \left(2+\sqrt{2}\right)^{2/3}+2
   \sqrt[3]{2+\sqrt{2}}-5=0$$
A: import numpy as np
import matplotlib.pyplot as plt


a = pow(2+np.sqrt(2),1/3) 
y, x = np.ogrid[-10:10:100j, -10:10:100j]
plt.contour(x.ravel(), y.ravel(), x**3-3*x*y+3*a, [6], colors='r')
plt.contour(x.ravel(), y.ravel(), y**3-3*a*x*y+3*a*a, [9])
plt.grid()
plt.show()


A: Write the equations as $$
x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}\\
\frac{(xy)^{3}}{x^3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}
$$
Now let $z = xy$ which gives you 
$$
f(z) = z^3 - (9 + 3z(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}})(6+ 3z - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}) = 0
$$
You can analyse $f(z)$ easily and find that  it has only one root at roughly $z = 14.485$. 
From here, the original two equations give $x$ and $y$.
A: $$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}}$$
$$y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}} $$
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].

