9.5 The numbers $x$, $y$, and $z$ satisfy the equations

$$
x y + y z + z x = x y z, \quad x + y + z = 1
$$

What values can the sum $x^{3} + y^{3} + z^{3}$ take?
Solution 1: Let $x y z=p$. Then, from the condition $x y+y z+z x$ is also equal to $p$. Therefore, by Vieta's theorem, the numbers $x, y$, and $z$ are the roots of the polynomial $t^{3}-t^{2}+p t-p$. However, the number 1 is a root of such a polynomial, so one of the numbers is equal to 1. Then the other two numbers are opposite, and the sum of the cubes of all three is 1.

Solution 2: From the condition $x y z=x y+y z+z x$. Then $x^{3}+y^{3}+z^{3}=3 x y z+(x+y+z)\left(x^{2}+y^{2}+\right.$ $\left.z^{2}-x y-y z-z x\right)=3(x y+y z+z x)+\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)=(x+y+z)^{2}=1$.

Solution 3: From the condition

$$
0=(x y z-x y-y z-z x)+(x+y+z-1)=(x-1)(y-1)(z-1) .
$$

Therefore, one of the numbers is equal to 1. Then the other two numbers are opposite, and the sum of the cubes of all three is 1.

## Criteria:

- If it is stated without proof that one of the numbers is equal to 1, and the answer to the problem is derived from this -1 point;
- If it is proven that one of the numbers is equal to 1, but the problem is not solved to the answer -6 points.