Solution
= Solution
Define
$$
\boxed{A=\frac{\mu_1(1-\beta_1)}{r_1^3}+\frac{\mu_2}{r_2^3}}.
$$
At an <equilibrium point>, the velocity and acceleration vanish. Direct differentiation gives
$$
U_z=-Az,
\qquad
U_y=(1-A)y,
$$
and
$$
U_x=(1-A)x-\mu_1\mu_2\left[(1-\beta_1)r_1^{-3}-r_2^{-3}\right].
$$
Thus the equilibrium conditions are
$$
\boxed{Az=0,
\qquad
(1-A)y=0,
\qquad
(1-A)x=\mu_1\mu_2\left[(1-\beta_1)r_1^{-3}-r_2^{-3}\right]}.
$$