1 | initial version |

This system is equivalent to the equation: $(x+y+z)^3 = 24xzy + 627$. By AM-GM, we have $$27xyz \leq (x+y+z)^3 = 24xzy + 627,$$ implying that $xyz \leq 209$. That is, there are only a finite number of solution, which is easy to find by brute-force.

2 | No.2 Revision |

This system is equivalent to the equation: $(x+y+z)^3 = 24xzy + 627$. By ~~AM-GM, ~~AM-GM inequality, we have
$$27xyz \leq (x+y+z)^3 = 24xzy + 627,$$
implying that $xyz \leq 209$. That is, there are only a finite number of solution, which is easy to find by brute-force.

3 | No.3 Revision |

This system is equivalent to the equation: $(x+y+z)^3 = 24xzy + 627$. By AM-GM inequality, we have
$$27xyz \leq (x+y+z)^3 = 24xzy + 627,$$
implying that $xyz \leq 209$. That is, there are only a finite number of solution, which ~~is ~~are easy to find by brute-force.

4 | No.4 Revision |

This system is equivalent to the equation: $(x+y+z)^3 = 24xzy + 627$. By AM-GM inequality, we have
$$27xyz \leq (x+y+z)^3 = 24xzy + 627,$$
implying that $xyz \leq 209$. That is, there are only a finite number of ~~solution, ~~solutions, which are easy to find by brute-force.

5 | No.5 Revision |

I assume that $x,y,z,n$ are nonnegative.

This system is equivalent to the equation: $(x+y+z)^3 = 24xzy + 627$. By AM-GM inequality, we have $$27xyz \leq (x+y+z)^3 = 24xzy + 627,$$ implying that $xyz \leq 209$. That is, there are only a finite number of solutions, which are easy to find by brute-force.

**ADDED.** One can also apply AM-GM in the other direction:
$$(x+y+z)^3 = 24xzy + 627 \leq \frac{24}{27}(x+y+z)^3 + 627,$$
implying that $(x+y+z)^3 \leq 5643$ and thus $x+y+z\leq 17$. This given $n=0$ as the only possible option, however it would mean $xyz=-25$, a contradiction.

6 | No.6 Revision |

I assume that $x,y,z,n$ are nonnegative.

This system is equivalent to the equation: $(x+y+z)^3 = 24xzy + 627$. By AM-GM inequality, we have $$27xyz \leq (x+y+z)^3 = 24xzy + 627,$$ implying that $xyz \leq 209$. That is, there are only a finite number of solutions, which are easy to find by brute-force.

**ADDED.** One can also apply AM-GM in the other direction:
$$(x+y+z)^3 = 24xzy + 627 \leq \frac{24}{27}(x+y+z)^3 + 627,$$
implying that $(x+y+z)^3 \leq 5643$ and thus $x+y+z\leq 17$. This ~~given ~~gives $n=0$ as the only possible option, however it would mean $xyz=-25$, a contradiction.

7 | No.7 Revision |

I assume that $x,y,z,n$ are nonnegative.

This system is equivalent to the equation: $(x+y+z)^3 = 24xzy + 627$. By AM-GM inequality, we have $$27xyz \leq (x+y+z)^3 = 24xzy + 627,$$ implying that $xyz \leq 209$. That is, there are only a finite number of solutions, which are easy to find by brute-force.

**ADDED.** One can also apply AM-GM in the other direction:
$$(x+y+z)^3 = 24xzy + 627 \leq \frac{24}{27}(x+y+z)^3 + 627,$$
implying that $(x+y+z)^3 \leq 5643$ and thus ~~$x+y+z\leq ~~$24n+3=x+y+z\leq 17$. This gives $n=0$ as the only possible option, however it would mean $xyz=-25$, a contradiction.

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