The dumbest solution for eraser cube puzzle

cube.py

from itertools import permutations

pieces = [[0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0], 
          [0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0], 
          [0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1], 
          [1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1], 
          [0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0], 
          [0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1]]

def rotations(piece):
    res = []
    for i in range(4):
        res += [piece[i*3:]+piece[:i*3]]
    return res

def match(pieces):
    # check 8 corners (4 bottom and 4 on top)
    if (pieces[0][0] + pieces[1][9] + pieces[4][6] != 1 or
        pieces[0][3] + pieces[2][9] + pieces[1][6] != 1 or
        pieces[0][6] + pieces[3][9] + pieces[2][6] != 1 or
        pieces[0][9] + pieces[4][9] + pieces[3][6] != 1 or
        pieces[5][9] + pieces[1][0] + pieces[4][3] != 1 or
        pieces[5][0] + pieces[4][0] + pieces[3][3] != 1 or
        pieces[5][3] + pieces[3][0] + pieces[2][3] != 1 or
        pieces[5][6] + pieces[2][0] + pieces[1][3] != 1):
        return False
    
    # check 12 edges (4 bottom, 4 on top and 4 sides), 2 cells for each
    if (pieces[0][1] + pieces[1][8] != 1 or
        pieces[0][2] + pieces[1][7] != 1 or
        pieces[0][4] + pieces[2][8] != 1 or
        pieces[0][5] + pieces[2][7] != 1 or
        pieces[0][7] + pieces[3][8] != 1 or
        pieces[0][8] + pieces[3][7] != 1 or
        pieces[0][10] + pieces[4][8] != 1 or
        pieces[0][11] + pieces[4][7] != 1 or
        
        pieces[5][8] + pieces[1][1] != 1 or
        pieces[5][7] + pieces[1][2] != 1 or
        pieces[5][5] + pieces[2][1] != 1 or
        pieces[5][4] + pieces[2][2] != 1 or
        pieces[5][2] + pieces[3][1] != 1 or
        pieces[5][1] + pieces[3][2] != 1 or
        pieces[5][11] + pieces[4][1] != 1 or
        pieces[5][10] + pieces[4][2] != 1 or
        
        pieces[1][4] + pieces[2][11] != 1 or
        pieces[1][5] + pieces[2][10] != 1 or
        pieces[2][4] + pieces[3][11] != 1 or
        pieces[2][5] + pieces[3][10] != 1 or
        pieces[3][4] + pieces[4][11] != 1 or
        pieces[3][5] + pieces[4][10] != 1 or
        pieces[4][4] + pieces[1][11] != 1 or
        pieces[4][5] + pieces[1][10] != 1):
      
        return False
    
    return True
  
def solve(pieces):
    solutions = [] # we're going to find all the possible solutions
    r0 = pieces[0]
    for r1 in rotations(pieces[1]):
        for r2 in rotations(pieces[2]):
            for r3 in rotations(pieces[3]):
                for r4 in rotations(pieces[4]):
                    for r5 in rotations(pieces[5]):
                        for perm in list(permutations([r1, r2, r3, r4, r5])):
                            if match([r0] + list(perm)):
                                solutions.append([r0] + list(perm))
    return solutions
  
solve(pieces)