11.2. Find all solutions in real numbers of the system of equations

$$
\left\{\begin{array}{l}
x(1+y z)=9 \\
y(1+x z)=12 \\
z(1+x y)=10
\end{array}\right.
$$
Answer: $x=1, y=4, z=2$.

Solution. Subtract the first equation from the second and third, we get: $y-x=3, z-x=1$. Substitute the expressions $y=x+3, z=x+1$ into the first equation, we get $x(x+3)(x+1)+x=9$, which after expanding the brackets leads to the cubic equation $x^{3}+4 x^{2}+4 x-9=0$. One of its roots is $x=1$, factorize the left side $x^{3}+4 x^{2}+4 x-9=(x-1)\left(x^{2}+5 x+9\right)$. The discriminant of the second bracket is negative, so the only real root of the equation $x^{3}+4 x^{2}+4 x-9=0$ and the solution to the original system is $x=1$. Then $y=x+3=4, z=x+1=2$.

Remark. After finding the real root $x=1$, its uniqueness can be proven in another way by investigating the function
$f(x)=x^{3}+4 x^{2}+4 x-9$. Its derivative $f(x)=3 x^{2}+8 x+4$ has roots $x_{1}=-2, x_{2}=-\frac{2}{3}$, it is greater than zero to the left of the first and to the right of the second of them, and negative between them. Therefore, the function $f(x)$ is increasing on the intervals $(-\infty,-2] \cup\left[-\frac{2}{3},+\infty\right)$ and decreasing on the interval $\left[-2,-\frac{2}{3}\right]$. Thus, the point $x_{1}=-2$ is a local maximum, and the point $x_{2}=-\frac{2}{3}$ is a local minimum. The values of the function at these points $f(-2)=-9, f\left(-\frac{2}{3}\right)=-\frac{275}{27}$ are negative, so its graph can only intersect the x-axis on the interval $\left(-\frac{2}{3},+\infty\right)$, on which it strictly increases. Therefore, the equation $f(x)=0$ cannot have more than one solution, the one we have already found, $x=1$.

Grading criteria. ($\cdot$) The answer is guessed and verified: 1 point.

($\cdot$) Expressions for all variables in terms of one, such as $y=x+3, z=x+1$: 1 point. ($\cdot$) The expressions are substituted into one of the equations and a cubic equation of the form $x^{3}+4 x^{2}+4 x-9=0$ is obtained: 1 point. ($\cdot$) A root of the cubic equation, such as $x=1$, is noticed and verified: 2 points. ($\cdot$) The values of the other variables are calculated from it: 1 point. ($\cdot$) It is proven that the cubic equation has no other roots and the solution is unique: 2 points.