Let $ f(a)=\frac{a^3+12a}{3a^2+4} $ where $ a \in \mathbb{R} $. And we have to find the root of the following equation: $ f(f(f(x)))=x $ Notice that $ f'(a)=\frac{3\left(a^2-4\right)^2}{\left(3a^2+4\right)^2} \ge 0, \forall a \in \mathbb{R} $. Therefore, if $ f(f(f(x)))=x $, then $ f(x)=x \Leftrightarrow x=\frac{x^3+12x}{3x^2+4} \Leftrightarrow x=-2,0,2 $ And since $ y=f(x)=x, z=f(f(x))=f(x)=x $, so the solutions are $ \{x,y,z\}=\{-2,-2,-2\},\{0,0,0\},\{2,2,2\}. \blacksquare $

ComplexPhi wrote: Find all the triplets of real numbers $(x , y , z)$ such that : $y=\frac{x^3+12x}{3x^2+4}$ , $z=\frac{y^3+12y}{3y^2+4}$ , $x=\frac{z^3+12z}{3z^2+4}$ Alternative much easy to understand,i think. If we multiply the realation we obtain $xyz = xyz\frac{x^2+12}{3x^2+4}\cdot\frac{y^2+12}{3y^2+4}\cdot\frac{z^2+12}{3z^2+4}$ or $\frac{x^2+12}{3x^2+4}\cdot\frac{y^2+12}{3y^2+4}\cdot\frac{z^2+12}{3z^2+4} = 1$. Now if $|x|,|y|,|z| > 2$ $\implies$ $x^2 > 4$ and $3x^2+4 > x^2+12$ $\implies$ $\frac{x^2+12}{3x^2+4}\cdot\frac{y^2+12}{3y^2+4}\cdot\frac{z^2+12}{3z^2+4} < 1$. If $|x|,|y|,|z| < 2$ $\implies$ $x^2 < 4$ and $3x^2+4 < x^2+12$ $\implies$ $\frac{x^2+12}{3x^2+4}\cdot\frac{y^2+12}{3y^2+4}\cdot\frac{z^2+12}{3z^2+4} > 1$. So $|x| = |y| = |z| = 2$ If $x = -2$ $\implies$ $y = z = -2$. If $x = 2$ $\implies$ $y = z = 2$. The solutions are $(x,y,z)\in$ $\{$$(2,2,2)$$,$$(-2,-2,-2)$$\}$.

The thing what I just understood is that ur solution is obviously wrong

BSJL wrote: The thing what I just understood is that ur solution is obviously wrong yeah you are right,sorry i didn't observe the case when |x| < 2 and |y|,|z| > 2

Let's see it from a different perspective more appropriate for Juniors (even if they are from Romania ). First suppose that $|x|,|y|,|z|\neq 2,0$.Now add $2$ in each relation to get $y+2=\frac{(x+2)^3}{3x^2+4}$ and other $2$ similar relations. Multiply the three relations and simplify to get $(x+2)^2(y+2)^2(z+2)^2=(3x^2+4)(3y^2+4)(3z^2+4) \ \bf \color{red} (1)$ Now subtract $2$ from each of the initial relations of the system to get $y-2=\frac{(x-2)^3}{3x^2+4}$ and other two similar relations. Multiplying all three relations and simplifying gives $(x-2)^2(y-2)^2(z-2)^2=(3x^2+4)(3y^2+4)(3z^2+4)$. Multiply this one with $\bf \color{red} (1)$ by parts to obtain $\prod_{sym} (x^2-4)^2=\prod_{sym} (3x^2+4)^2 \ \bf \color{blue} (2)$ However it's easy to see that $(3x^2+4)^2>(x^2-4)^2 \ \forall x\neq 0$.Similar relations hold for $y,z$. Since all the terms of these relations are positive,we can multiply them to get $\prod_{sym} (x^2-4)^2<\prod_{cyc} (3x^2+4)^2$ which contradicts $\bf \color{blue} (2)$ Thus,no solutions satisfy the initial restrictions whence we get that at least one among $x,y,z$ equals $0$ or $2$. Case work gives $(x,y,z)=(0,0,0),(2,2,2),(-2,-2,-2)$ as the only triples that satisfy.

ComplexPhi wrote: Find all the triplets of real numbers $(x , y , z)$ such that : $y=\frac{x^3+12x}{3x^2+4}$ , $z=\frac{y^3+12y}{3y^2+4}$ , $x=\frac{z^3+12z}{3z^2+4}$

Very food solution for juniors,reeh_haan!Thanks .Also,gavrilos!

First of all, if we rewrite the equations as polynomial equations, we find they are all cubic. So including multiplicities, there are $3^3=27$ solutions over $\mathbb{C}$. Let $x=2X$, $y=2Y$, $z=2Z$ We get $Y=\frac{X^3+3X}{3X^2+1}$, $Z=\frac{Y^3+3Y}{3Y^2+1}$, $X=\frac{Z^3+3Z}{3Z^2+1}$ From this we see three obvious real solutions for $(x,y,z)$, which are $(0,0,0)$, $(2,2,2)$, and $(-2,-2,-2)$. We now show the other $24$ solutions are not in real numbers by finding them. Consider the trigonometric identity $\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$. This is a special case of the more general $\tan n\theta=\frac{\binom{n}{1}\tan\theta-\binom{n}{3}\tan^3\theta+\binom{n}{5}\tan^5\theta-\cdots}{\binom{n}{0}-\binom{n}{2}\tan^2\theta+\binom{n}{4}\tan^4\theta-\cdots}$. So if $X=i\tan\theta$, we have $Y=i\tan 3\theta$, $Z=i\tan 9\theta$, and $X=i\tan 27\theta$ Thus $27\theta=\theta+n\pi\implies \theta=\frac{n\pi}{26}$ with $n\in\mathbb{Z}$. So the other $24$ tuples $(x,y,z)$ are $\left(2i\tan\frac{n\pi}{26},2i\tan\frac{3n\pi}{26},2i\tan\frac{9n\pi}{26}\right)$ for $1\leq n\leq 25$, $n\neq 13$.

rchokler wrote: First of all, if we rewrite the equations as polynomial equations, we find they are all cubic. So including multiplicities, there are $3^3=27$ solutions over $\mathbb{C}$. Let $x=2X$, $y=2Y$, $z=2Z$ We get $Y=\frac{X^3+3X}{3X^2+1}$, $Z=\frac{Y^3+3Y}{3Y^2+1}$, $X=\frac{Z^3+3Z}{3Z^2+1}$ From this we see three obvious real solutions for $(x,y,z)$, which are $(0,0,0)$, $(2,2,2)$, and $(-2,-2,-2)$. We now show the other $24$ solutions are not in real numbers by finding them. Consider the trigonometric identity $\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$. This is a special case of the more general $\tan n\theta=\frac{\binom{n}{1}\tan\theta-\binom{n}{3}\tan^3\theta+\binom{n}{5}\tan^5\theta-\cdots}{\binom{n}{0}-\binom{n}{2}\tan^2\theta+\binom{n}{4}\tan^4\theta-\cdots}$. So if $X=i\tan\theta$, we have $Y=i\tan 3\theta$, $Z=i\tan 9\theta$, and $X=i\tan 27\theta$ Thus $27\theta=\theta+n\pi\implies \theta=\frac{n\pi}{26}$ with $n\in\mathbb{Z}$. So the other $24$ tuples $(x,y,z)$ are $\left(2i\tan\frac{n\pi}{26},2i\tan\frac{3n\pi}{26},2i\tan\frac{9n\pi}{26}\right)$ for $1\leq n\leq 25$, $n\neq 13$. Splendid ideas!

Let $f(u)=\frac{u^3+12u}{3u^2+4}$. Then, note that $$ \frac{f(u)+2}{f(u)-2}=\frac{(u+2)^3}{(u-2)^3}. $$Thus, if we set $a=g(x)$, $b=g(y)$ and $c=g(z)$ for $g(v)=\frac{v+2}{v-2}$, then we get the equalities $a=c^3$, $b=a^3$ and $c=b^3$. Now it's easy to finish the solution. P. S. It's just a sketch of a solution. In particular, one also need to work out the case when some of $x,y,z$ equals 2.