Example 2 Let $x_{1}, x_{2}, x_{3}$ be the three roots of the equation $x^{3}+x+2=0$. Then the determinant
$$
D=\left|\begin{array}{lll}
x_{1} & x_{2} & x_{3} \\
x_{2} & x_{3} & x_{1} \\
x_{3} & x_{1} & x_{2}
\end{array}\right|=(\quad \text {. }
$$
(A) -4
(B) -1
(C) 0
(D) 2
From Vieta's formulas, we have $x_{1}+x_{2}+x_{3}=0$.
By the definition of the determinant, we know
$$
D=3 x_{1} x_{2} x_{3}-\left(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}\right)=0 .
$$