Example 6 Find the real solutions of the equation $\left(1+x+x^{2}\right)(1+x+\cdots$ $\left.+x^{10}\right)=\left(1+x+\cdots+x^{6}\right)^{2}$.
(54th Moscow Mathematical Olympiad)
Solution: Clearly, $x=0$ is a solution to the original equation.
When $x \neq 0$, we have
$$
\begin{array}{c}
\frac{1+x+\cdots+x^{10}}{1+x+\cdots+x^{6}}=\frac{1+x+\cdots+x^{6}}{1+x+x^{2}}, \\
\text { i.e., } \frac{1+x+\cdots+x^{6}}{1+x+x^{2}}=\frac{x^{7}+x^{8}+x^{9}+x^{10}}{x^{3}+x^{4}+x^{5}+x^{6}}=x^{4},
\end{array}
$$

Thus, $1+x+\cdots+x^{6}=x^{4}+x^{5}+x^{6}$,
$$
\begin{array}{l}
1+x+x^{2}+x^{3}=0, \\
(1+x)\left(1+x^{2}\right)=0,
\end{array}
$$

Therefore, $x=-1$.
Hence, the real solutions to the original equation are $x=0$ or $x=-1$.