Example 1. Let $x$ be a cube root of 1 different from 1, find the value of $x^{\text {D}}$ $+x^{2}$. $(n \in N)$
Solve: From $x^{3}=1$, i.e., $x^{3}-1=0$, which is also $(x-1)\left(x^{2}+x+1\right)=0$.
Since $x \neq 1$, it follows that $x^{2}+x=-1$.
When $n=3k$,
$$
x^{\mathrm{n}}+x^{2 n}=\left(x^{3}\right)^{k}+\left(x^{3}\right)^{2 k}=2 \text {; }
$$

When $n=3 k+1$,
$$
\begin{array}{l}
x^{\mathrm{n}}+x^{2 \mathrm{n}}=\left(x^{3}\right)^{k} \cdot x+\left(x^{3}\right)^{2 k} \cdot x^{2} \\
=x+x^{2}=-1 ;
\end{array}
$$

When $n=3 k+2$,
$$
\begin{array}{l}
x^{n}+x^{20} \\
=\left(x^{3}\right)^{1} \cdot x^{2}+\left(x^{3}\right)^{2 k} \cdot x^{3} \cdot x \\
=x^{2}+x=-1
\end{array}
$$