(ab)−1ab=e(ab)^{-1}ab=e(ab)−1ab=e, so (ab)−1a=b−1(ab)^{-1}a=b^{-1}(ab)−1a=b−1 and (ab)−1=b−1a−1(ab)^{-1}=b^{-1}a^{-1}(ab)−1=b−1a−1
Answer: (ab)−1=b−1a−1(ab)^{-1}=b^{-1}a^{-1}(ab)−1=b−1a−1
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