How prove this equation have more one solution $x_{1}+x_{2}+\cdots+x_{n}=d,x_{1}x_{2}\cdots x_{n}=b$

Question

Let $x_{i}\in \mathbb{Z},i=1,2,\ldots,n$, and such that $1\le x_{1}\le x_{2}\le\cdots\le x_{n}$. Show that $$\begin{cases} x_{1}+x_{2}+\cdots+x_{n}=d\\ x_{1}x_{2}\cdots x_{n}=b\\ b,d\in \mathbb{Z} \end{cases}$$ has at most one solution $(x_{1},x_{2},\cdots,x_{n})$.

Answer 1

$n=4$, $d=12$, $b=48$. Solutions $(1,3,4,4)$ and $(2,2,2,6)$.