Find all real numbers $x, y, z$ satisfying:

$$
\left\{\begin{aligned}
(x+1) y z & =12 \\
(y+1) z x & =4 \\
(z+1) x y & =4
\end{aligned}\right.
$$

Proposition 1.27 (Vieta's Formulas). Let $P(x)=a x^{2}+b x+c$ be a real polynomial of degree 2 (with $a \neq 0$) having $z_{1}$ and $z_{2}$ as real roots. Then $z_{1} z_{2}=\frac{c}{a}$ and $z_{1}+z_{2}=-\frac{b}{a}$.

Proof. According to Corollary 1.15, we have $P(x)=a\left(x-z_{1}\right)\left(x-z_{2}\right)$. By expanding the right-hand term, we find the announced equalities.

Remark 1.28. These relations are useful because they express the coefficients of the polynomial in terms of the roots. For this reason, the following exercise will be sought.
Let $(x, y, z)$ be a solution. Clearly, none of these numbers is zero. By subtracting the third equation from the second equation, we deduce that $z x=x y$, and by simplifying by $x$ (which is non-zero), we obtain that $z=y$. By subtracting the third equation from the first equation, we get: $y^{2}-x y=8$, or equivalently $x y=y^{2}-8$. The second equation can be rewritten as $y^{2} x+x y=4$. Therefore, we have:

$$
y\left(y^{2}-8\right)+y^{2}-8=4
$$

or equivalently $y^{3}+y^{2}-8 y-12=0$. We notice that $y=3$ is a solution. By performing the Euclidean division of $y^{3}+y^{2}-8 y+12$ by $y-3$, we find:

$$
y^{3}+y^{2}-8 y-12=(y-3)\left(y^{2}+4 y+4\right)=(y-3)(y+2)^{2}
$$

We deduce that $y=z=3$ or $y=z=-2$. In the first case, $x=\frac{1}{3}$ and in the second case, $x=2$. Conversely, the triplets $(2,-2,-2)$ and $\left(\frac{1}{3}, 3,3\right)$ are solutions and these are the only ones.