This is one possible solution path. It will be useful to produce a list of the possible values for the letters in clues, derived from the 23 squares from 33² = 1089 to 99² = 9801. The numbers given by their last one to three digits are, in ascending order with duplicates removed, {1, 4, 5, 6, 9, 16, 21, 25, 49, 56, 61, 64, 69, 84, 89, 96, 100, 184, 216, 249, 296, 304, 356, 521, 561, 569, 600, 601, 625, 649, 761, 764, 801, 916, 969}. We don’t need to worry about leading zeros because any such number, like 1089 → 089, will duplicate the value of the shorter 1089 → 89. Zero itself is specifically excluded by the preamble.

For 6ac, h³ is a two-digit cube, of which there are only two: 3³ = 27 and 4³ = 64; but 3 can’t be the last digit of a square, so h = 4 and 6ac = 64. From 8ac = I³ + 2L with three digits, I ≤ 9. From 17ac = a³ + C, which is < 7dn = 4__ (from the grid), a < 8 and must be one of {5, 6} (4 is already taken by h and 1 would make 10dn = a³ + aI a maximum of 10). 19ac = 16 + I² + a has three digits and 16 + a ≤ 22, so I² ≥ 100 − 22 = 78, ie I ≥ 9, which makes I = 9 and 29ac = 27. Now 19ac = 97 + a is one of {102, 103} and we can enter the first two digits in the grid, along with 1 as the first digit of 16dn and 28dn, which are < 19dn which starts with 1. Also, 10dn = a³ + 9a is one of {170, 270} and we can enter the 70.

8ac is 729 + 2L and 5dn = 2L² + 4 is between that and 1000, which requires L² − L > 362.5 and also L² + 2 < 500, so L < √498 ≈ 22.3. The only integers in that range are {20, 21, 22}, of which only 21 is in the list of allowed values, so L = 21, 8ac = 771 and 5dn = 886. 1ac and 1dn are clued between these two, so they must start with 7 or 8. Now 3dn = 16S + 21 is between 886 and 1000, and from the grid its middle digit is 1, so it must start with 91. Therefore S is between (911 − 21)/16 ≈ 55.6 and (919 − 21)/16 ≈ 56.1, ie S = 56 and 3dn = 917. Also 25dn > 3dn from the clue order, so it must start with 9.

For 2dn = m − 21 we have _7_ in the grid; since it’s < 20dn, which starts with 2 or 3 (from the last digit of 19ac), 2dn starts with one of {1, 2, 3}. So m is in one of the ranges [191, 200], [291, 300] or [391, 400]; from the list of allowed values, the only one that fits is m = 296, making 2dn = 275. This limits the first digit of 17ac, 31ac and 9ac to {2, 3} and the first digit of 24ac and 23dn to {1, 2}. For 24ac = [12]9_ to be < 23dn = [12]1_, they must be 19_ and 21_ respectively.

From the grid, we have [23][12]8 for 9ac = 144 + u, so u ends in 4 and is in the range from 218 − 144 = 74 to 328 − 144 = 184. The allowed values in that range are {84, 184}, so 9ac is one of {228, 328}, either of which makes 10dn = 270 and a = 6. That resolves 19ac = 103, making 20dn start with 3; since 17ac < 20dn < 17dn and both 17 entries must start with the same digit, all of {17ac, 31ac, 27ac, 17dn} start with 3. We’ve now completed 9ac = 328, so u = 184. 17ac = 216 + C is 3_0 in the grid, so C ends in 4 and is in the range [84, 174]; the only such value is C = 84, making 17ac = 300. Now 19dn = 93 + N for 1_2 in the grid, so N ends in 9 and is in the range [19, 99] (not 9 because that’s I, and because 19dn > 16dn = 109), so N is one of {49, 69, 89}.

For 20dn = U − 21 we have 33_ in the grid, so 352 ≤ U ≤ 360; the only possible value is U = 356, making 20dn = 335. Now 30ac starts with 5, so 4dn, 12ac, 4ac and 22dn must all start with 4 or 5. 4ac = F + 2N + 184 is 4_8 or 5_8 in the grid; because N ends in 9, F must end in 6. F is ≥ 408 − 184 − 2×89 = 46 and ≤ 598 − 184 − 2×49 = 316. Since 56 and 296 are already taken, F is one of {96, 216}. For 13ac = 6 + b + 2F we have 7_7, so b ends in 9. For 14ac = 126 + b we have 6__ in the grid, which puts b in the range [474, 573]; the only matching value is b = 569, making 14ac = 695. Now 13ac = 575 + 2F = 7_7, so F = 96 and 13ac = 767.

For 4dn = 2c + 4 we have 436 or 536 in the grid, so c is one of {216, 266}; but 266 isn’t an allowed value, so c = 216 and 4dn = 436. 12ac = iN + 6 is 45_, so i is between (450 − 6)/89 ≈ 4.99 and (459 − 6)/49 ≈ 9.2; the only available value is i = 5. Now N is between (450 − 6)/5 = 88.8 and (459 − 6)/5 = 90.6, ie N = 89, making 12ac = 451, 4ac = 458 and 19dn = 182.

For 11dn = 10×O we have _9_ in the grid and from the clue order it starts with 5 or 6, so it’s one of {590, 690}; but 59 isn’t an allowed value, so O = 69 and 11dn = 690.

For 1ac = 579 + p we have 72_ or 82_ in the grid, so p is in the range [141, 150] or [241, 250]. The only available value is p = 249, with 1ac = 828.

For 28ac = 5n + 21 the answer is between 103 and 182, so n is between (105 − 21)/5 = 16.8 and (181 − 21)/5 = 32, which can only be n = 25, with 28ac = 146. Now 30ac = 46 + y = 5__ means y is in the range [454, 553] and the only available value is y = 521, which makes 30ac = 567, 22dn = 527 and 26dn = 602. 24ac = r − e − 4 is now completed as 196, so r − e = 200; the remaining pairs that are 200 apart are (561, 761) and (601, 801). For 25dn = 6(r − f) we have 96_ in the grid; to be a multiple of 6 it has to be one of {960, 966}, but 966 would give 362 for 31ac, which is supposed to be between 17ac = 300 and 9ac = 328; therefore 26dn = 960 and r − f = 160. The options for f = r − 160 are {761 − 160 = 601, 801 − 160 = 641}, of which only f = 601 is valid, with r = 761 and e = 561.

For 7dn = 25(1 + R) we have 4_5, which, to be a multiple of 25, is either 425 or 475, so R is in {16, 18}; but 18 isn’t a valid value, so R = 16, which makes 7dn = 425 and 16dn = 109. 21ac is _52 in the grid and in the clue order it’s between 695 and 767, so it can only be 21ac = 752. Now 17dn = 16t − 601 is 37_ in the grid, which only allows t = 61, making 17dn = 375. From the grid we now have 27ac = 254 + 2s = 352, so s = 49. Similarly, 21ac = 552 + 2o = 752, so o = 100, which makes 23dn = 218. Also, 31ac = 366 − T = 302, so T = 64. Finally, 1dn = E + 120 = 8_4, between 828 and 886, so E ends in 4 and is in the range [714, 764]. The only valid value in that range is E = 764, making 1dn = 884.

Now we have all the letter values, and in ascending order the letters are [h i a I R L n s S t T O C N F o u c p m U y e b f r E]. The method for extracting the message is “related to the title”, so it must have something to do with threes. Taking letters three places apart, we can get “hIntCopyf”, “iRsTNumer” and “aLSOFcUbE”, which, with a bit of respacing and case adjustment, can be read as “hint: copy first numerals of cube”. This suggests a rule for the sequence of raising a number to the power of three and then taking the first three digits as the next number. Trying this on the known grid entries confirms that in almost every case the result is another entry.

For example, the cube of 1ac = 828 is 567663552 and its first three digits are 567, which is 30ac. Then 567³ = 182284263, which leads to 182 = 19dn, and so on. Chaining these together, we can get (302, 275, 207, 886, 695, 335, 375, 527, 146, 311, 300, 270, 196, 752, 425, 767, 451, 917, 771, 458, 960, 884, 690, 328, 352, 436, 828, 567, 182, 602, 218, 103, 109, 129, ...), where all but 207, 311 and 129 are entries in the grid. We can fit 207 at 18ac and 311 at 15ac, but 129 can’t go anywhere, so that must be the end of the partial sequence. The starting number 302 needs to be highlighted at 31ac.

Finally, the sum of the three-digit entries is 15625 = 5⁶ and the sum of the letter values is 5832 = 2³×3⁶. To express them in a “form that relates to the rule for the sequence”, they need to be recognised as cubes, namely 25³ and 18³ respectively.