Solve the following system of equations:

$$
\begin{aligned}
x+y+z & =1 \\
x^{3}+y^{3}+z^{3} & =1 \\
x y z & =-16 .
\end{aligned}
$$
I. Solution: Raise both sides of (1) to the third power:

$$
x^{3}+y^{3}+z^{3}+3\left(x^{2} y+x y^{2}+x^{2} z+x z^{2}+y^{2} z+y z^{2}+2 x y z\right)=1
$$

Substitute the appropriate values from (2) and (3) into (4), we get

$$
x y(x+y)+x z(x+z)+y z(y+z)=32
$$

The binomials in the three parentheses can be expressed from (1) as follows:

$$
x y(1-z)+x z(1-y)+y z(1-x)=32
$$

From (3), we have

$$
x y+x z+y z=-16
$$

Based on these, we can write a cubic equation (let the unknown be $t$), whose roots are the sought values of $x, y$, and $z$. This equation is:

$$
(t-x)(t-y)(t-z)=t^{3}-(x+y+z) t^{2}+(x y+x z+y z) t-x y z=0
$$

Using the equalities (1), (5), and (3),

$$
t^{3}-t^{2}-16 t+16=0
$$

(6) can also be written as:

$$
t^{2}(t-1)-16(t-1)=(t-1)\left(t^{2}-16\right)=(t-1)(t-4)(t+4)=0
$$

Thus, the roots of (6) are

$$
t_{1}=1, \quad t_{2}=4, \quad t_{3}=-4
$$

Since our system of equations in $x, y, z$ is completely symmetric, any permutation of the three values we obtained is a set of roots, so the values of $x, y, z$ are respectively: $1,4,-4 ; 1,-4,4 ; 4,1,-4 ; 4,-4,1 ;-4,1,4 ;-4,4,1$.

Gergely Ervin (Bp. IV., Konyves Kalman g. IV. o. t.)

Note: If we eliminate two unknowns from (5) using (1) and (3), we naturally obtain essentially the same equation (6).

Gereben Ildiko (Bp. XXI., Jedlik g. III. o. t.)

II. Solution: (4) can be written as follows, considering (1):

$$
x^{2} y+x y^{2}+x^{2} z+x z^{2}+y^{2} z+y z^{2}+2 x y z=0
$$

The left side can be transformed into a product as follows:

$$
\begin{aligned}
& x^{2} y+x y^{2}+x^{2} z+x y z+x z^{2}+y z^{2}+y^{2} z+x y z= \\
= & x y(x+y)+x z(x+y)+z^{2}(x+y)+y z(x+y)= \\
= & (x+y)\left(x y+y z+x z+z^{2}\right)=(x+y)[y(x+z)+z(x+z)]= \\
= & (x+y)(x+z)(y+z) .
\end{aligned}
$$

Thus, our equation can also be written as

$$
(x+y)(y+z)(z+x)=0
$$

From this, either

a) $x+y=0$, and then from (1) and (3),

$z_{1}=1, \quad x_{1}=4, \quad y=-4$

$z_{2}=1, \quad x_{2}=-4, \quad y_{2}=4 ; \quad$ or

b) $y+z=0$, then from (1) and (3),

$x_{3}=1, \quad y_{3}=4, \quad z_{3}=1$

and so on. Clearly, we obtain the same sets of roots as in the first solution.

Gati Gyula (Debrecen, Vegyip. t. IV. o. t.)