Given that,
∣a∣=2n+1 and aba−1=b−1|a|=2^{n+1} \ and \ aba^{-1}=b^{-1}∣a∣=2n+1 and aba−1=b−1
we have to find b2=?b^2=?b2=?
since aba−1=b−1 ⟹ (aba−1)−1=(b−1)−1aba^{-1}=b^{-1} \implies (aba^{-1})^{-1}=(b^{-1})^{-1}aba−1=b−1⟹(aba−1)−1=(b−1)−1
⟹ ab−1a−1=b\implies ab^{-1}a^{-1}=b⟹ab−1a−1=b
⟹ (ab−1a−1)2=b2\implies (ab^{-1}a^{-1})^2=b^2⟹(ab−1a−1)2=b2
⟹ b2=ab−1a−1ab−1a−1\implies b^2=ab^{-1}a^{-1}ab^{-1}a^{-1}⟹b2=ab−1a−1ab−1a−1 =ab−2a−1=ab^{-2}a^{-1}=ab−2a−1
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