I have to solve the system of equations $$\begin{cases} x+y+z=2008,\\ xyz=14000, \end{cases}$$ where $x,y,z$ are positive integers such that $1\le x \le y \le z \le 2000.$
My work so far:
Let $P(t)=(t-2000)(t-7)(t-1)$ and $Q(t)=(t-x)(t-y)(t-z)$.
Then $P(t)-Q(t)=kt$. So $2000-z\ge 0 \Rightarrow Q(2000)\ge 0$
$14000=2^4\cdot 5^3\cdot 7$ has $40$ divisors, but just $36$ of them are $\leq 2000$ and just $6$ of them are between $\frac{2008}{3}$ and $2000$, namely $\{700,875,1000,1400,1750,2000\}$. These values are the possible values for $z$. For instance, $z=2000$ leads to the solution $x=1,y=7$. $z=1750$ does not lead to integer solutions, neither they do $1400,1000,875$ or $700$. So $\color{red}{(1,7,2000)}$ is the only solution.