question_answer 17)

 that         = (a ? b) (b ? c) (c ? a) (a + b + c)  

Answer:

Let       Operate  we get             Taking (b ? a) and (c ? a) as common from C2 and C3 respectively.                   Expand along R1, we get                   = (b ? a) (c ? a)       [c2 + ac + a2 ? b2 ? ab ? a2]       = (b ? a) (c ? a) (c2 ? b2 + ac ? ac)       = (b ? a) (c ? a) ((c ? b) (c + b)                         + a (c ? b))       = (b ? a) (c ? a) (c + b + a)        = (a ? b) (b ? c) (c ? b) (a + b + c)