Solve the following system of equations:

$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x+y+z=\frac{13}{2} ; \quad x y z=1
$$
I. solution: Our system of equations can be written as:

$$
\begin{array}{r}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{13}{3} \\
x+y+z=\frac{13}{3} \\
x y z=1
\end{array}
$$

From (1) and (3), it follows that none of the three unknowns can be 0, so we can multiply (1) by $x y z$ and (2) by $x$. We obtain that

$$
\begin{gathered}
y z+x z+x y=\frac{13}{3} x y z=\frac{13}{3} \\
x^{2}+x y+x z=\frac{13}{3} x
\end{gathered}
$$

Subtracting (4) from (5)

$$
x^{2}-y z=\frac{13}{3} x-\frac{13}{3}
$$

Substituting $\frac{1}{x}$ for $y z$ based on (3), multiplying by $x$, and rearranging

$$
x^{3}-\frac{13}{3} x^{2}+\frac{13}{3} x-1=0
$$

The left side can be transformed as follows:

$$
\begin{gathered}
x^{3}-1-\frac{13}{3} x(x-1)=(x-1)\left(x^{2}+x+1\right)-\frac{13}{3} x(x-1)= \\
=(x-1)\left(x^{2}-\frac{10}{3} x+1\right)=0
\end{gathered}
$$

from which

$$
x_{1}=1
$$

The two roots of the quadratic factor equation are

$$
x_{2}=3, \quad x_{3}=\frac{1}{3}
$$

Substituting these values of $x$ into (2) and (3), we obtain the complete system of roots:

| $x=1$ | 1 | 3 | 3 | $\frac{1}{3}$ | $\frac{1}{3}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $y=3$ | $\frac{1}{3}$ | 1 | $\frac{1}{3}$ | 1 | 3 |
| $\mathrm{z}=\frac{1}{3}$ | 3 | $\frac{1}{3}$ | 1 | 3 | 1 |

Károly Bartók (Székesfehérvár, József Attila g. II. o. t.)

II. solution: The factored form of the cubic equation is

$$
a\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)=0
$$

Setting $a-1$, then

$$
x^{3}-\left(x_{1}+x_{2}+x_{3}\right) x+\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right) x-x_{1} x_{2} x_{3}=0
$$

Considering equations (2), (3), and (4), the 3 roots of our system of equations are thus the same as the 3 roots of the following cubic equation.

$$
x^{3}-\frac{13}{3} x+\frac{13}{3} x-1=0
$$

The 3 roots obtained and their $3!=6$ permutations give the 6 triplets of roots of our system of equations. äsmallskip