I'm very weak in math and logic, and currently tried doing K-map, and got this as result: $$(C'B')+(CB)$$
My question is, can this be further simplified?
I tried it myself, but I got $0$ (False).
$(C'C)+(B'B)$ I just swapped the places of B' and C. I don't know if can do that.
$0+0$
$0$
No you cannot just swap characters like you did, and the simplification is certainly not $0$.
What you start with is simplified, and it is expressed as a sum of products.
Alternatively, we can write the expression as a product of sums:
$$\begin{align} (C'B') + (CB) &= (C'+CB)(B'+CB) \\\\ & = \underbrace{(C'+C)}_{\large =\, 1}(C'+B)(B'+C)\underbrace{(B'+B)}_{\large = \,1} \\\\&= (C'+B)(B'+C)\end{align}$$
What we have is equivalent to the biconditional (which can be read as "B, if and only if C", and which in logic can be expressed $$(C\rightarrow B) \land (B\rightarrow C) = (B\leftrightarrow C)$$
Note that the given expression is equivalent to $(B\oplus C)'$ which reads (NOT (B XOR C)):
$$\begin{align} (B\oplus C)' &= [(BC')+(B'C)]' = (BC')'(B'C)' \\ \\ &=(B'+C)(B+C')\end{align}$$ This is, in the end, equivalent to the product of sums form (which in turn is equivalent to the sum of products form of the expression.