Regular hexagons can be put together in a flower-like way:
From the picture above you can see that hexagons are necessary to build the hexagon-flower on the left, and hexagons are necessary to build the hexagon-flower on the right.
Neither nor are perfect cubes. As we add more hexagons to make larger hexagon-flowers, will we ever be able to make a hexagon-flower whose total amount of hexagons is a perfect cube ?
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The number of regular hexagons used follows the sequence a n = 3 n ( n + 1 ) + 1 = ( n + 1 ) 3 − n 3 . See the solution in the Inspiration link for the proof.
Suppose a n can be expressed as a (positive) perfect cube, then this would contradict Fermat's last theorem , which is absurd!