Tackling this problem requires a comically deep look into how dominoes work. Any chunk of 2x1 pieces can be expressed as a chain of dominoes lined head-to-tail.
Notice that the dominoes will always go from black-to white and it's impossible to form a chain of them where you don't end up with a checker pattern because within a domino: the black is adjacent to the white and between them, you've got black adjacent to white and the pattern continues meaning it's impossible to have two whites or two blacks be adjacent to eachother. The second thing to realise is that they occupy an even amount of spaces because you've got an integral number of 2x1's. The third thing is that they'll end on a square which is a different colour than at the start of the chain, again due to the nature of how they are even and how one end has a tail and one end has a head. Finally: because there is an integral number of dominoes and each domino has a black and a white there is an equal number of blacks and whites.
So we know that you can't fit dominoes into an odd number of squares but the mutilated grid doesn't have a problem with that. The problem is that you can't make a checkerboard with an equal number of blacks and whites with the mutilated grid and therefore you cannot fill it with dominoes because if dominoes can't be placed in a checkerboard formation with an equal number of black and white squares, you will not be able to make them fit.
So there you go. This has other applications like if you had just cancelled out two random squares in the original 8x8 grid. If you can form a chain of dominoes, you know it's possible and if you have an equal number of white and black squares with no blatant obstructions, you can form a chain.
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