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Real Solution of the system of equations

$x^3-6z^2+12z-8 = 0$

$y^3-6x^2+12x-8 = 0$

$z^3-6y^2+12y-8 = 0$

$\bf{My\; Try::}$ We can write the given equations as

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

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

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

Now I did not understand how can i solve after that

Help Required , Thanks

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By letting $x = y = z = t$ you get $(t-2)^3 = 0$ it $(x,y,z) = (2,2,2)$ is a solution. Is it unique? I don't know and wouldn't think so. – user88595 Apr 11 '14 at 13:00

Hawk's answer shows that, if any of the numbers is not 2, then at least one number is more than 2. Suppose $z>2$. Then $6(z-1)^2+2>8$. How does that relate to $x$, and then how does that relate to $y$?

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Hint: Add all the terms and you will get $(x-2)^3+(y-2)^3+(z-2)^3=0$.

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@BarryCipra Yes, you are correct...these typos occur so much! I will rectify them! – Hawk Apr 11 '14 at 13:38

For $x>2$,

$\begin{align} -6x(x-2)&<&0 &<& (x-2)^3\\ -6x^2+12x&<&0 &<& x^3-6x^2+12x-8\\ 8&<&6x^2-12x+8 &<& x^3\\ 2&<&(6x^2-12x+8)^{1/3} &<& x \end{align}$

Similarly for $0<x<2$,

$\begin{align*} x<(6x^2-12x+8)^{1/3} &< 2 \end{align*}$

And if $x\le0$,

$\begin{align*} 2\le(6x^2-12x+8)^{1/3} \end{align*}$

Let $f(x) = (6x^2-12x+8)^{1/3}$

Thus for the system of equations above to hold, we have

$x=f(z), y=f(x), z=f(y)$

Hence $z=f(f(f(z))$

But if $z > 2$, $2 < f(f(f(z))) < f(f(z)) < f(z) < z$ so $f(f(f(z))) \ne z$

And if $0 < z < 2$, $z < f(z) < f(f(z)) < f(f(f(z))) < 2$ so $f(f(f(z))) \ne z$

And if $z \le 0$, $2 \le f(f(f(z))) \le f(f(z)) \le f(z)$ so $f(f(f(z))) \ne z$

hence the only solution is $x=y=z=2$

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