Find all complex numbers $m$ such that polynomial
\[x^3 + y^3 + z^3 + mxyz\]
can be represented as the product of three linear trinomials.
1. **Express the polynomial in terms of linear trinomials:**
   We start with the polynomial 
   \[
   P(x, y, z) = x^3 + y^3 + z^3 + mxyz
   \]
   and assume it can be factored into the product of three linear trinomials:
   \[
   P(x, y, z) = \prod_{i=1}^3 (\alpha_i x + \beta_i y + \gamma_i z + \delta_i)
   \]
   By homogeneity, we can set \(\delta_1 = \delta_2 = \delta_3 = 0\). This simplifies our polynomial to:
   \[
   P(x, y, z) = \prod_{i=1}^3 (\alpha_i x + \beta_i y + \gamma_i z)
   \]

2. **Assume non-zero coefficients and rescale:**
   Assume all coefficients \(\alpha_i, \beta_i, \gamma_i\) are non-zero. Without loss of generality, we can rescale the coefficients such that \(\alpha_1 = \alpha_2 = \alpha_3 = 1\). This gives us:
   \[
   P(x, y, z) = (x + \beta_1 y + \gamma_1 z)(x + \beta_2 y + \gamma_2 z)(x + \beta_3 y + \gamma_3 z)
   \]

3. **Compare coefficients for \(P(x, y, 0)\):**
   Consider the polynomial when \(z = 0\):
   \[
   P(x, y, 0) = x^3 + y^3 = (x + \beta_1 y)(x + \beta_2 y)(x + \beta_3 y)
   \]
   By expanding and comparing coefficients, we get:
   \[
   \beta_1 + \beta_2 + \beta_3 = 0
   \]
   \[
   \beta_1 \beta_2 + \beta_2 \beta_3 + \beta_3 \beta_1 = 0
   \]
   \[
   \beta_1 \beta_2 \beta_3 = 1
   \]
   These are the roots of the polynomial \(t^3 - 1 = 0\), so \(\beta_1, \beta_2, \beta_3\) are the cube roots of unity:
   \[
   \{\beta_1, \beta_2, \beta_3\} = \{1, \varepsilon, \varepsilon^2\}
   \]
   where \(\varepsilon = e^{2\pi i / 3}\).

4. **Analogous reasoning for \(\gamma_i\):**
   Similarly, for the \(\gamma_i\) coefficients, we have:
   \[
   \{\gamma_1, \gamma_2, \gamma_3\} = \{1, \varepsilon, \varepsilon^2\}
   \]

5. **Determine valid factorizations:**
   We now try the six permutations of the \(\gamma_i\) values to find valid factorizations. The valid factorizations are:
   \[
   (x + y + z)(x + \varepsilon y + \varepsilon^2 z)(x + \varepsilon^2 y + \varepsilon z)
   \]
   \[
   (x + y + \varepsilon z)(x + \varepsilon y + z)(x + \varepsilon^2 y + \varepsilon^2 z)
   \]
   \[
   (x + y + \varepsilon^2 z)(x + \varepsilon y + \varepsilon z)(x + \varepsilon^2 y + z)
   \]

6. **Calculate \(m\) for each factorization:**
   For each factorization, we expand and compare with the original polynomial \(x^3 + y^3 + z^3 + mxyz\):
   \[
   (x + y + z)(x + \varepsilon y + \varepsilon^2 z)(x + \varepsilon^2 y + \varepsilon z) = x^3 + y^3 + z^3 - 3xyz
   \]
   \[
   (x + y + \varepsilon z)(x + \varepsilon y + z)(x + \varepsilon^2 y + \varepsilon^2 z) = x^3 + y^3 + z^3 - 3\varepsilon xyz
   \]
   \[
   (x + y + \varepsilon^2 z)(x + \varepsilon y + \varepsilon z)(x + \varepsilon^2 y + z) = x^3 + y^3 + z^3 - 3\varepsilon^2 xyz
   \]
   Thus, the values of \(m\) are:
   \[
   m \in \{-3, -3\varepsilon, -3\varepsilon^2\}
   \]

The final answer is \(\boxed{m \in \{-3, -3\varepsilon, -3\varepsilon^2\}}\)