Answer
By the inductive hypothesis
we can fill each of these eight objects with tiles. Putting these tilings together produces the desired tiling.
Work Step by Step
--showing that a three-dimensional 2^n × 2^n × 2^n checkerboard
with one 1 × 1 × 1 cube missing can be completely
covered by 2 × 2 × 2 cubes with one 1 × 1 × 1 cube removed.
--Let P(n) be the statement that every 2^n×2^n×2^n checkerboard
with a 1 × 1 × 1 cube removed can be covered by tiles
that are 2×2×2 cubes each with a 1×1×1 cube removed.
-- The basic step, P(1), holds because one tile coincides with the
solid to be tiled. Now assume that P(k) holds. Now consider a
2k+1×2k+1×2k+1 cube with a 1×1×1 cube removed.
-Split this object into eight pieces using planes parallel to its faces
and running through its center. The missing 1 × 1 × 1 piece
occurs in one of these eight pieces. Now position one tile with
its center at the center of the large object so that the missing
1 × 1 × 1 cube lies in the octant in which the large object is
missing a 1×1×1 cube.
-This creates eight 2k×2k×2k cubes,
each missing a 1 × 1 × 1 cube. By the inductive hypothesis
we can fill each of these eight objects with tiles. Putting these
tilings together produces the desired tiling.
