Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$

Question

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?

Answer 1

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 $$\frac{x^6}{27}-\frac{2}{3} \left(1+\sqrt[3]{2+\sqrt{2}}\right) x^3+\frac{-6+\sqrt{2}+12 \sqrt[3]{2+\sqrt{2}}-6 \left(2+\sqrt{2}\right)^{2/3}}{x^3}-2 \left(2+\sqrt{2}\right)^{2/3}+2 \sqrt[3]{2+\sqrt{2}}-5=0$$

Answer 2

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()

Answer 3

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 $$x^3=44.9381694189876 \implies x= ??? \implies y= ???$$

Answer 4

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 $y$.

Answer 5

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

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].