Consider truth table of three variable XOR function
A |
B |
C |
A⊕B⊕C |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 (m1) |
0 |
1 |
0 |
1(m2) |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
1 (m4) |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 (m7) |
Our XOR function is also known as Odd function because it outputs 1 whenever we have odd number of 1's in the input.
The minterms corresponding to the A⊕B⊕C is m1,m2,m4,m7 as we can see from truth table.
So, f(A⊕B⊕C)= ∑ m(1,2,4,7) = (A'B'C+ A'BC' + AB'C' + ABC) (Equivalent expression expressed as Sum of products)
So clearly A⊕B⊕C will be an SOP expression having 3 literals in each minterm.
Options (a) and (b) are ruled out as they cannot generate a 3 literal minterm(such as ABC) which can justify that they correspond to A⊕B⊕C.
Now if we expand option (c)
ABC + A'(B⊕C)+B'(A⊕C)
ABC + A'(B'C+BC') + B'(A'C + AC')
ABC (m7) + A'BC' (m2) + A'B'C (m1) + AB'C' (m4)
So above terms correspond exactly to our previously derived result of A⊕B⊕C.
Hence Option (c) is answer.