[题目]-|||-证明:-|||-x -1 0 0-|||-0 x -1 0 =(a)_(3)(x)^3+(a)_(2)(x)^2+(a)_(1)x+(a)_(0)-|||-0 0 x -1-|||-a0 a1 a2 a3

【解析】
步骤 1：按第1列展开行列式
\[
\begin{vmatrix}
x & -1 & 0 & 0 \\
0 & x & -1 & 0 \\
0 & 0 & x & -1 \\
a_0 & a_1 & a_2 & a_3
\end{vmatrix} = x \begin{vmatrix}
x & -1 & 0 \\
0 & x & -1 \\
a_1 & a_2 & a_3
\end{vmatrix} - a_0 \begin{vmatrix}
-1 & 0 & 0 \\
x & -1 & 0 \\
0 & x & -1
\end{vmatrix}
\]

步骤 2：计算第一个3阶行列式
\[
\begin{vmatrix}
x & -1 & 0 \\
0 & x & -1 \\
a_1 & a_2 & a_3
\end{vmatrix} = x \begin{vmatrix}
x & -1 \\
a_2 & a_3
\end{vmatrix} - (-1) \begin{vmatrix}
0 & -1 \\
a_1 & a_3
\end{vmatrix} = x(xa_3 + a_2) + a_1 = x^2a_3 + xa_2 + a_1
\]

步骤 3：计算第二个3阶行列式
\[
\begin{vmatrix}
-1 & 0 & 0 \\
x & -1 & 0 \\
0 & x & -1
\end{vmatrix} = -1 \begin{vmatrix}
-1 & 0 \\
x & -1
\end{vmatrix} = -1((-1)(-1) - 0) = -1
\]

步骤 4：将步骤2和步骤3的结果代入步骤1
\[
x(x^2a_3 + xa_2 + a_1) - a_0(-1) = x^3a_3 + x^2a_2 + xa_1 + a_0
\]