A new December and a new bunch of puzzles from mscroggs.co.uk.

n = 99999…999

= 10⁵⁵ – 1

n³ = (10⁵⁵ – 1)³

= (10⁵⁵)³ + 3*(10⁵⁵)²*(-1) + 3*(10⁵⁵)*(-1)² + (-1)³

= 10¹⁶⁵ – 3*10¹¹⁰ + 3*10⁵⁵ – 1

 10000...0000...0000...000
-         300...0000...000
+                300...000
-                        1
==========================
  9999...9700...0299...999

So now we just have to count the 9s and 7s and the 2.

The 1 lives in position 1. The 3 above it has a 3 in position 56. The next 3 is in position 111. Finally the 1 above it is in position 166.

The result of adding and subtracting is 9s from position 1 to 55, a 2 at position 56, then some 0s, a 7 at position 111 and finally 9s from position 112 to 165. 55*9+2+7+54*9 = 990. Which happens to be 55*18.

Just to test my logic here: if n had been 9999 (4 digits), n³ would be 999,700,029,999. 4*9+2+7+3*9 = 72 = 4*18. Checks out.