## Task 2 - 271242

Determine all triples $(x, y, z)$ of real numbers that satisfy the following system of equations (1), (2), (3):

$$
\begin{aligned}
& 1 \cdot x^{3}+9 \cdot y^{2}+8 \cdot y+8=1988 \\
& 1 \cdot y^{3}+9 \cdot z^{2}+8 \cdot z+8=1988 \\
& 1 \cdot z^{3}+9 \cdot x^{2}+8 \cdot x+8=1988
\end{aligned}
$$
It is obvious that the three equations can be transformed into each other by cyclically swapping $x, y$, and $z$. Therefore, a triplet with $x=y=z$ would automatically be a solution if $x$ satisfies the equation

$$
x^{3}+9 x^{2}+8 x=1980
$$

An equally obvious solution is $x=10$, so the triplet $(x, y, z)=(10,10,10)$ is already a solution. We will now show that there can be no further solutions in $\mathbb{R}$. We perform polynomial division of equation (4) by $x-10$, and obtain

$$
x^{2}+19 x+198=0
$$

This quadratic equation has no solutions in $\mathbb{R}$. Therefore, any further solution triplets would require that the three variables are pairwise distinct. We now consider the two functions

$$
f(x)=x^{3}
$$

and

$$
g(y)=9 y^{2}+8 y+8
$$

The latter is an upward-opening parabola, with its minimum at $y=-\frac{4}{9}$. (Any conclusion about the properties of one variable applies equally to all three variables due to cyclic symmetry.) Both (6) for $y \geq-\frac{4}{9}$ and (5) generally are monotonically increasing.

We now assume that all three variables $x, y, z \geq-\frac{4}{9}$. In equation (1), this means that due to monotonicity, starting from the triplet $(10,10,10)$, if $x$ is set greater than 10, then $y$ must be less than 10. If $y$ is less than 10, then equation (2) implies that $z$ must be greater than 10. Thus, both $x$ and $z$ are each greater than 10, making equation (3) unfulfillable and leading to a contradiction. The same applies in the opposite direction. We therefore investigate whether a solution is possible if (at least) one of the variables is less than $-\frac{4}{9}$. It is easy to show by setting the first derivative of (6) to zero and substituting that

$$
9 y^{2}+8 y+8 \geq \frac{56}{9}
$$

Using this in equation (1), we get:

$$
\begin{gathered}
9 y^{2}+8 y+8=1988-x^{3} \geq \frac{56}{9} \\
x \leq \sqrt[3]{1988-\frac{56}{9}}=a
\end{gathered}
$$

Thus, there is an upper bound

$$
\text { (7) } \quad x, y, z \leq a
$$

with $a=\sqrt[3]{1981 \frac{7}{9}} \approx 12.56$. Now, assume that $x \leq-\frac{4}{9}$. From equation (1), it follows:

$$
\text { (8) } 9 y^{2}+8 y=1980-x^{3} \geq 1980+\left(\frac{4}{9}\right)^{3}
$$

Since $y=a$ does not satisfy this inequality, either $y>a$ must hold, which contradicts (7), or $y$ must be less than the other, negative solution of the quadratic equation. That is:

$$
y \leq-\frac{4}{9}-\sqrt{\frac{16}{81}+\frac{1980-x^{3}}{9}}
$$

We claim that

$$
-\frac{4}{9}-\sqrt{\frac{16}{81}+\frac{1980-x^{3}}{9}} < -\frac{4}{9}
$$

which is indeed the case for all $x$, as shown above. q.e.d. The only solution triplet is thus $(x, y, z)=(10,10,10)$.