<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] 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()


</s>