Altogether you have 6 bars of each color, meaning you require at least 10 connections of blue-blue and black-black (5 each). Each corner does 2 connections if you have all 3 of the same color, 1 connection if you have 2 of one and 1 of other color, and the chewed off part has 0 connections. So, at minimum, you require 3 corners with all 3 of the same color (4 with 1/2, and the chewed off corner is arbitrary).
Now, can we make 3 corners with all 3 rods the same color? Let's say the bottom right corner (the one with 3x blue on the image), has all colors the same. It is obviously impossible that either of the 3 corners next to it has 3 of the same color, as it would lead to having face with 2x same color connectors. So, 3 corners with all the same color connections need to be for example 3 of (0, 0, 0), (1, 1, 0), (0, 1, 1) and (1, 0, 1). Or 3 of the other 4 corners. Sounds great, we can have the thingy connected.
There is a tiny little issue that makes everything impossible. If we have say 000 corner all blue, then 110, 011 and 101 can only be all black to not have the same color on the face. 2 faces even. But if then 110 is all black, 011 or 101 cannot be. So, we can have just 2 corners all of the same color, but we require 3 to satisfy the connectivity.