R(ABCDE)
A-->BC,CD->E B->D E->A
candidate keys are {A,E,CD,CB}
so partial dependency in B->D....first decomposition will be BD and ABCE
now BD is fine but in ABCE candiate key is E...so due to A->BC there is again partial dependency...
second decomposition will be of ABCE...it will be into ABC and AE..now everything is in BCNF also
final answer :- 3 decompositions are requires...BD,ABC,AE