A.not(B)+AC

abc + aB + aBc
ac(b+B) + aB
ac x 0 + aB
aB

B = your Bold B

Not sure if this is correct.

I have made little bit clear

The confusion here is, it seems it is giving to different answers

A.B.C +A.Bnot

Another is

A.C+A.Bnot

I am little bit confused here. Does it has two solution?

 

Strictly following the precedence i.e (. before +)
A.B.C  + A. B (not A . not C) (De morgans law)
A.B.C + A.B.notA.notC
A.B.C + (A.notA).B.notC

A and not A = False

ABC + False

so if any of the A or B or C is false the result is False and for the expression to be true they all have to be True

I am assuming that the bnot.(A+C) = b.not(A+C)

Please consult the solution below, the above is solved on wrong assumption :D

assuming bnot = not b

a.b.c + a.notb.a + a.notb.c (distributive)
 
a.b.c + a.notb + a.notbc ( a.a = a)

a.c.b   + a.c.notb  +a.notb (re-arranging)

ac(b+ notb) + a.notb ( distributive)

b+ notb = true

therefore ac + a.notb