Three. (20 points) Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>$ $0, b>0)$, $A$ and $B$ are two points on the same branch of the hyperbola. The perpendicular bisector of segment $AB$ intersects the $x$-axis at point $P$. Find the range of the $x$-coordinate of point $P$.
Let $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and $P\left(x_{0}, 0\right)$, with the midpoint of $AB$ being $M\left(x^{\prime}, y^{\prime}\right)$. Then,
$$
\frac{x_{1}^{2}}{a^{2}}-\frac{y_{1}^{2}}{b^{2}}=1, \frac{x_{2}^{2}}{a^{2}}-\frac{y_{2}^{2}}{b^{2}}=1 \text {. }
$$

Subtracting the two equations gives
$$
\frac{x_{1}^{2}-x_{2}^{2}}{a^{2}}-\frac{y_{1}^{2}-y_{2}^{2}}{b^{2}}=0 \text {. }
$$

Thus, the slope of $AB$ is
$$
k_{A B}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{b^{2}\left(x_{1}+x_{2}\right)}{a^{2}\left(y_{1}+y_{2}\right)}=\frac{b^{2} x^{\prime}}{a^{2} y^{\prime}} \text {. }
$$

Therefore, the slope of the perpendicular bisector $l$ of segment $AB$ is
$$
k_{l}=-\frac{1}{k_{A B}}=-\frac{a^{2} y^{\prime}}{b^{2} x^{\prime}} \text {. }
$$

Hence, the equation of $l$ is
$$
y-y^{\prime}=-\frac{a^{2} y^{\prime}}{b^{2} x^{\prime}}\left(x-x^{\prime}\right) \text {. }
$$

Setting $y=0$, we get
$$
a^{2} y^{\prime} x_{0}=\left(a^{2}+b^{2}\right) y^{\prime} x^{\prime} \text {. }
$$

If $y^{\prime}=0$, by the fact that $A$ and $B$ are points on the same branch of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, $A$ and $B$ must be symmetric with respect to the $x$-axis, and the perpendicular bisector of $AB$ would coincide with the $x$-axis, which contradicts the problem statement. Therefore, $y^{\prime} \neq 0$. Thus,
$$
x_{0}=\frac{a^{2}+b^{2}}{a^{2}} \cdot x^{\prime} \text {. }
$$

Since $x_{1}$ and $x_{2}$ have the same sign, and $\left|x_{1}\right| \geqslant a$, $\left|x_{2}\right| \geqslant a$ (with equality holding for at most one of them), it follows that $\left|x^{\prime}\right|>a$.
Thus, $x_{0}>\frac{a^{2}+b^{2}}{a}$ or $x_{0}<-\frac{a^{2}+b^{2}}{a}$.