23. (CZS 4) Find all complex numbers $m$ such that polynomial
$$
x^{3}+y^{3}+z^{3}+\text { mxyz }
$$
can be represented as the product of three linear trinomials.
23. We may assume w.lo.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 $x^{3}+y^{3}+(-a x-b y)^{3}+m x y(-a x-b y)=\left(1-a^{3}\right) x^{3}-(3 a b+$ $m)(a x+b y) x y+\left(1-b^{3}\right) y^{3} \equiv 0$. This is obviously equivalent to $a^{3}=b^{3}=1$ and $m=-3 a b$, from which it follows that $m \in\left\{-3,-3 \omega,-3 \omega^{2}\right\}$, 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.