Answer
${r_e} =\sqrt[6] \frac{2A}{B} $
Work Step by Step
Let $r_e$ = equilibrium seperation
$F = - \frac{dU}{dr}_{_{r=r_e}} =0$
$ \frac{dU}{dr}_{_{r=r_e}} =-0$
$\frac{d}{dr}(\frac{A}{r^{12}} - \frac{B}{r^6})=0$
$\frac{d}{dr}(\frac{A}{r^{12}} - \frac{B}{r^6})= 0$
$\frac{6B}{r^{7}_e}-\frac{12A}{r^{16}_e} =0$
$\frac{1}{r^{7}_e}({6B}-\frac{12A}{r^{6}_e}) =0$
$({6B}-\frac{12A}{r^{6}_e}) =0 \times {r^{7}_e}$
$({6B}-\frac{12A}{r^{6}_e}) =0 $
Multiply the equation by ${r^{6}_e}$:
$({6B}{r^{6}_e}-{12A}) =0 \times {r^{6}_e} $
$({6B}{r^{6}_e}-{12A}) =0 $
${6B}{r^{6}_e} ={12A}) $
${r^{6}_e} =\frac{12A}{6B} $
${r^{6}_e} =\frac{2A}{B} $
${r_e} =\sqrt[6] \frac{2A}{B} $
