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Question Number 122484 by JBocanegra last updated on 17/Nov/20
4a3b⋅213⋅a23⋅b13(43⋅213)(aa23⋅b43)(223⋅213)(a1−23⋅b−43)=(22−13)(a13⋅b−43)⋅13(3233)(a3⋅1b43)=323⋅13⋅a3.b83⋅1b4=32⋅a⋅b33⋅b4
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