Reducing A'B+(ABC)'+C(B'+A) with boolean algebra

Question

what is the possible way to reduce this equation with Boolean algebra to smallest number of terms

A'B+(ABC)'+C(B'+A)

Answer 1

first of all try to ask question in more organized way second, make sure you know elements of boolean algebra this can be reduced with DeMorgan rule

apply DeMorgan rule

A'B=A+B'
(ABC)'=A'+B'+C'
C(B'+A)=C'+(A+B')'

then

A'B+(ABC)'+C(B'+A)=(A+B')+A'+B'+C'+C'+(A+B')'

let x=(A+B')

A'B+(ABC)'+C(B'+A)=X+A'+B'+C'+C'+X'
X + X' =1

finally

A'B+(ABC)'+C(B'+A)=A'+B'+C'+C'=A'+B'+C'

final result is

A'+B'+C'