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Here candidate keys are AB ,BC and F.

There is partial dependency in FD ( B->D ).

I think C->A is not partial dependency ,since A is key attribute.

So to eliminate that divide the table into two tables as

take B+ = { B,D,E } as one table and remaining as others and B is the common attribute .

So (ABCF) , (BDE) .

Correct if anything wrong.
edited by

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