Under E1 conditions (typically hot, polar protic solvent), one of the bromines will leave as bromide to give a carbocation. Ignoring the isotope effect, there's about an equal propensity for either of the isotopic bromines to leave, giving about a 50/50 ratio of the two carbocations. Deprotonation at the beta-position gives an alkene. I've only shown the most substituted possible alkene since that will be the most stable and major product.
(Note, the carbocation could rearrange, but I'm ignoring that for this level of analysis.)
Under more strongly basic conditions an E2 reaction is more likely. There is a strong preference for an anti-periplanar arrangement of the leaving group, carbons where the alkene will form, and the deprotonated hydrogen. This stereochemical requirement is more easily visualized in the chair forms, which are not equivalent. The structure on the left is more stable since the methyl group is placed equatorially. In this conformation, there are two hydrogens that are anti-periplanar with the bromine-81 atom (green). A bulky base would react with the least hindered hydrogen (purple) and eliminate the bromine-81 to give the disubstituted alkene shown. A smaller base would be more likely to produce the most stable product, which results from deprotonating the hydrogen in blue.
Just for completeness I've shown the elimination product of the less stable chair conformation. In this case, there is only one anti-periplanar arrangement. If the orange hydrogen is deprotonated, then the bromine-79 atom can be eliminated to give the cyclohexene shown.