Question

Y^3 - 6x^2 + 12x - 8 = 0, z^3 - 6y^2 + 12y - 8 =0,x^3 - 6z^2 + 12z - 8 =0

Answer 1

Сразу видно что это суммы двучлена в кубе если суммировать все эти уравнение одновременно! 
$y^3-6x^2+12x-8=0\\ z^3-6y^2+12y-8=0\\ x^3-6z^2+12z-8=0\\ \\ y^3-6x^2+12x-8+z^3-6y^2+12y-8+x^3-6z^2+12z-8=0\\ (y^3-6y^2+12y-8)+(z^3-6z^2+12z-8 )+ (x^3-6x^2+12x-8)=0\\ (y-2)^3+(z-2)^3+(x-2)^3=0$
отудого следует что решение x=y=z=2

Answer 2

The given system of equations consists of three cubic equations:

1. $Y^3 - 6x^2 + 12x - 8 = 0$
2. $z^3 - 6y^2 + 12y - 8 = 0$
3. $x^3 - 6z^2 + 12z - 8 = 0$

To solve this system, we can start by isolating one variable in each equation and then substituting the expressions into the other equations. Let's solve for $Y$ in the first equation:

$Y^3 = 6x^2 - 12x + 8$

Now, let's solve for $z$ in the second equation:

$z^3 = 6y^2 - 12y + 8$

Finally, let's solve for $x$ in the third equation:

$x^3 = 6z^2 - 12z + 8$

Now, we can substitute these expressions into the respective equations to obtain a system of three equations in one variable:

1. $(6x^2 - 12x + 8)^3 - 6x^2 + 12x - 8 = 0$
2. $(6y^2 - 12y + 8)^3 - 6y^2 + 12y - 8 = 0$
3. $(6z^2 - 12z + 8)^3 - 6z^2 + 12z - 8 = 0$

Solving this system of equations may be computationally intensive. If you have any specific values for $x$, $y$, or $z$ that you would like to substitute into the equations, I can assist you further.