You can not use associativity when operators are mixed:
$$a'.b'.c' + a.b'.c' + a.b.c' \neq (b'.c').(a'+ a) + a.b.c'\tag{1}$$
It is true that $a'.b'.c' = b'.c'.a'$, by commutativity of ".", but it is not legitimate/not valid to impose parentheses to group $(a' + a)$ as you did.
You can use the distributive laws, and you might want to use Demorgan's Laws, as well.
Example: for using the distributive law, and the fact that $b' + b = 1$
$$a.b'.c' + a.b.c' = a.c'.b' + a.c'.b = a.c'(b'+b) = a.c'$$
Now we've simplified our expression to
$$a'.b'.c' + a.c'\tag{2}$$
There's a common term of $c'$ in each of these products, so we can simplify further. See what you come up with, and I'll work with you to clarify/confirm, etc. if you have any more questions.
$$a'.b'.c' + a.c'=(a'.b' + a).c' $$ $$= [(a'+a).(b'+a)].c' $$ $$= (b'+a).c' = b'.c' + a.c' $$ $$= a.c' + b'.c'$$
Best Answer
Lets say X = (A + B)' * C') and Y = ((A' + B') * C')'
We know that XOR can be represented as
$X ⊕ Y = X\overline Y + \overline X Y$
So, first simplify the inners and once you have the simplified version, you can deal with the outer A.