Show that the body-centered-cubic crystal have three families of slip systems, i.e. twelve slip systems of (110)[111]-type, twelve slip systems of (112)[111]-type and twenty four slip systems of (123)[111]-type with a total of 48 slip systems.
What is the question here?
It is straightforward to show that the slip systems you mention exist, simply by considering the number of possible permutations of the indices. For the {110} systems, for example:
(110)[1-10] (101)[10-1] (011)[01-1]
(110)[-110] (101)[-101] (011)[0-11]
(1-10)[110] (10-1)[101] (01-1)[011]
(1-10)[-1-10] (10-1)[-10-1] (01-1)[0-1-1]
Note that six of these are simply the other six running in the reverse direction. Whether you consider there to be twelve {110} systems or six is a matter of convention.
(Note also that while these systems may be physically distinct, only five of them can be truly independent. For proof of this statement, see Kelly & Knowles, 'Crystallography and Crystal Defects'.)
If you are asking whether all 48 systems can be activated in any given bcc material, that is a very different question. It can only be answered by looking for papers on the material in question.
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