You have a 3x3x3 cube, built with 1x1x1 bricks (27 bricks in total).

[Werg22]

You have a 3x3x3 cube, built with 1x1x1 bricks (27 bricks in total). You remove two diagonally opposite bricks (long diagonals). Can you build the remaining figure from 2xx1x1 bricks?

 

[phlegmy]

i'm going to say:

Spoiler

no
the shapes volume is now 25 bricks
this would take 25 1x1x1 bricks
or 12.5 2x1x1 bricks
unless your allowed to have half a 2x1x1 brick, you can't do it?
unless I've missed something

 

[davee123]

Yeah, this problem becomes trivial if you've got an odd-dimension cube. However, it may still be impossible if it's a 4x4x4 cube, for different reasons, which I think is what the OP was going for. Just have to imply a different meaning of "Long Diagonal"-- that is, instead, you want to remove 2 cubes diagonally opposite on the same "plane" of the cube.

DaveE

 

[Werg22]

Dave is right, I didn't realize the triviality of an odd-order cube

Still though, like he pointed out, the 4x4x4 case, or any order for that matter, is impossible too. But the solution (at least the one I know) is much harder!

 

[sandy85]

There is a problem in dimension with its first digit. So, i am not guessing answer about this bricks game.