+ Let $n$ be a given integer greater than 5, solve the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+\cdots+x_{n}=n+2, \\
x_{1}+2 x_{2}+3 x_{3}+\cdots+n x_{n}=2 n+2, \\
x_{1}+2^{2} x_{2}+3^{2} x_{3}+\cdots+n^{2} \cdot x_{n}=n^{2}+n+4, \\
x_{1}+2^{3} x_{2}+3^{3} x_{3}+\cdots+n^{3} \cdot x_{n}=n^{3}+n+8,
\end{array}\right.
$$

where $x_{i} \geqslant 0, i=1,2, \cdots, n$.
Solution 1
$$
\begin{array}{l}
\sum_{k=1}^{n} k^{3} \cdot x_{k} \\
=\sum_{k=1}^{n-1}(k+1)(k-1)(k-2) \cdot x_{k}+2 \sum_{k=1}^{n-1} k^{2} x_{k}+\sum_{k=1}^{n} k x_{k}-2 \sum_{k=1}^{n} x_{k} \\
\leqslant n \sum_{k=1}^{n}(k-1)(k-2) \cdot x_{k}+2 \sum_{k=1}^{n} k^{2} \cdot x_{k}+\sum_{k}^{n} k \cdot x_{k}-2 \sum_{k}^{n} \cdot x_{k} \\
=(n+2) \sum_{k=:}^{n} k^{2} x_{k}+(1-3 n) \sum_{k}^{n-1} k x_{k}+(2 n-2) \sum_{k}^{n} \cdot x_{k} \\
=(n+2)\left(n^{2}-n+4\right)+(1-3 n) \cdot 2 n+(2 n-2)(n+1) \\
=n^{3}-3 n^{2}+4 n+6 \text {. } \\
\end{array}
$$
(Equality condition: $x_{3}=x_{4}=\cdots=x_{n}=0$ )
(Special case 1) Substituting)
$$
\begin{array}{l}
=(n+2)\left(n^{2}-n+4\right)+(1-3 n) \cdot 2 n+(2 n-2)(n+1) \\
=n^{3}-3 n^{2}+4 n+6 .
\end{array}
$$

Combining (1), we know that the inequality should hold with equality.
Solving $\left\{\begin{array}{l}x_{1}+x_{2}+x_{n} 1=n+1, \\ x_{1}+2 x_{2}+(n-1) x_{n} 1=2 n, \\ x_{1}+2^{2} x_{2}+(n-1)^{2} x_{n} 1=n^{2}-n+4,\end{array} \quad\left\{\begin{array}{l}x_{1}=n-1, \\ x_{2}=1 . \\ x_{n} 1=1 .\end{array}\right.\right.$
In summary: It is known that the equation has a unique solution $\left(x_{1}, x_{2}, x_{3}, x_{4}, \cdots, x_{n}\right.$, $\left.x_{n} 1\right)=(n-1,1,0,0, \cdots, 0,1)$.
$$
x_{1}+x_{2}+x_{3}+\cdots+x_{n}=n+2 \text {, }
$$

Solution 2 $\left\{\begin{array}{l}x_{1}+2 x_{2}+3 x_{3}+\cdots+n x_{n}=2 n+2, \\ x_{1}+2^{2} x_{2}+3^{2} x_{3}+\cdots+n^{2} x_{n}=n^{2}+n+4 . \\ x_{1}+2^{3} x_{2}+3^{3} x_{3}+\cdots+n^{3} x_{n}=n^{3}+n+8 .\end{array}\right.$
(2) - (1), (3) - (2), (1) - (3) yields
$$
\left\{\begin{array}{l}
x_{2}+2 x_{3}+\cdots+(n-1) x_{n}=n . \\
2 x_{2}+2 \cdot 3 x_{3}+\cdots+n(n-1) x_{n}=n^{2}-n+2 . \\
4 x_{2}+3^{2} \cdot 2 x_{3}+\cdots+n^{2}(n-1) x_{n}=n^{3}-n^{2}+4 .
\end{array}\right.
$$
(1i) $-2 \times$ (3). (7) $-2 \times$ (6) yields
$$
\left\{\begin{array}{l}
\sum_{i=1}^{n-1}(i-1)(i-2) x_{i}=n^{2}-3 n+2, \\
\sum_{i}^{n} i(i-1)(i-2) x_{i}=n^{3}-3 n^{2}+2 n .
\end{array}\right.
$$
(8) $\times n-$ (9) yields
$$
\sum_{i}^{n}(n-i)(i-1)(i-2) x_{i}=0, x_{i} \geqslant 0 .
$$

Therefore
$$
x_{i}=0(i=3,4, \cdots, n-1) .
$$

Substituting into (1), (2), (3) yields
$$
x_{1}=n, x_{2}=x_{n}=1 .
$$