# Revision history [back]

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.

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.

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.

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.

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.

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.

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.