23. (CZS 4) Find all complex numbers $m$ such that polynomial  
$$ x^{3}+y^{3}+z^{3}+m x y z $$  
can be represented as the product of three linear trinomials.
23. We may assume w.l.o.g. that in all the factors the coefficient of $x$ is 1. Suppose that $x + a y + b z$ is one of the linear factors of $p(x, y, z) = x^3 + y^3 + z^3 + m x y z$. Then $p(x)$ is 0 at every point $(x, y, z)$ with $z = -a x - b y$. Hence $x^3 + y^3 + (-a x - b y)^3 + m x y (-a x - b y) = (1 - a^3) x^3 - (3 a b + m)(a x + b y) x y + (1 - b^3) y^3 \equiv 0$. This is obviously equivalent to $a^3 = b^3 = 1$ and $m = -3 a b$, from which it follows that $m \in \{-3, -3 \omega, -3 \omega^2\}$, where $\omega = \frac{-1 + i \sqrt{3}}{2}$. Conversely, for each of the three possible values for $m$ there are exactly three possibilities $(a, b)$. Hence $-3, -3 \omega, -3 \omega^2$ are the desired values.