Example 5 Solve the equation $\sqrt{x-1}+\sqrt[3]{2-x}=1$.

Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. 

However, since the text provided is already in English, here is the equivalent text, maintaining the format:

Example 5 Solve the equation $\sqrt{x-1}+\sqrt[3]{2-x}=1$.
Solution: Since $\sqrt{x-1}, \frac{1}{2}, \sqrt[3]{2-x}$ form an arithmetic sequence, we set
$$
\left\{\begin{array}{l}
\sqrt{x-1}=\frac{1}{2}-d \\
\sqrt[3]{2-x}=\frac{1}{2}+d .
\end{array}\right.
$$
$(1)^{2}+(2)^{3}$, we get
$$
1=\left(\frac{1}{2}-d\right)^{2}+\left(\frac{1}{2}+d\right)^{3} \text {. }
$$

Simplifying and factoring, we get
$$
\begin{array}{l}
(2 d-1)(2 d+1)(2 d+5)=0 . \\
\therefore d_{1}=\frac{1}{2}, d_{2}=-\frac{1}{2}, d_{3}=-\frac{5}{2} .
\end{array}
$$

Substituting into (1), we get
$$
\sqrt{x-1}=0, \sqrt{x-1}=1, \sqrt{x-1}=3 \text {. }
$$

Thus, $x_{1}=1, x_{2}=2, x_{3}=10$.
Upon verification, the solutions to the original equation are $x_{1}=1, x_{2}=2$, $x_{3}=10$.

In general, irrational equations of the form $\sqrt[n]{f(x)} \pm \sqrt[m]{g(x)}=a$ (a constant), where $m, n \in N$, and $m, n>1, f(x) \pm g(x)$ is a constant, can be solved using the method of setting the common difference.