Example 4 Solve the equation $\sqrt{x-1}+\sqrt[3]{2-x}=1$.

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Solve the equation $\sqrt{x-1}+\sqrt[3]{2-x}=1$ by constructing an arithmetic sequence $\sqrt{x-1}, \frac{1}{2}, \sqrt[3]{2-x}$. Therefore, we set
$$
\left\{\begin{array}{l}
\sqrt{x-1}=\frac{1}{2}-d, \\
\sqrt[3]{2-x}=\frac{1}{2}+d .
\end{array}\right.
$$

By $(1)^{2}+(2)^{3}$, we get $1=\left(\frac{1}{2}-d\right)^{2}+\left(\frac{1}{2}+d\right)^{3}$.
Simplifying and factoring, we get $(2 d-1)(2 d+1)(2 d+5)=0$. Therefore, $d=\frac{1}{2}$ or $-\frac{1}{2}$ or $-\frac{5}{2}$. Substituting these values into (1), we get $\sqrt{x-1}=0$ or 1 or 3. Solving these, we get $x=1$ or 2 or 10. Upon verification, the solutions to the original equation are $x=1$ or 2 or 10.