This is one possible solving path.
It may be helpful to compile a complete list of the 45 cubes between 1 and 100,000 (from 2^{3} = 8 to 46^{3} = 97336), but we’ll minimise use of such a list here, concentrating on limits for the cube roots instead.
It’s useful to note that numbers ending in {2, 3, 7, 8} have cubes that end in {8, 7, 3, 2} respectively, and cubes of other numbers have the same last digit as the number itself.

Clue 7dn gives Y^{3} and as Y is a cube, say Y = *n*^{3}, we need *n*^{9} to have 5 digits.
That is, *n* is between 10000^{1/9} ≈ 2.8 and 99999^{1/9} ≈ 3.6, so *n* = 3, **Y = 27** and **7dn = 19683**.
Now 7ac = 2*b* has to fit 1_ in the grid, so *b* < 9.5 and the only available cube is ** b = 8**, making

For 16dn we have *i* + 64 = 5___, so *i* is in the range 5000 − 64 = 4936 to 5999 − 64 = 5935; if *i* = *n*^{3} then *n* is between ∛4936 ≈ 17.03 and ∛5935 ≈ 18.1, so *n* = 18, ** i = 5832** and

6dn = 1000 + L must have four digits.
For 9dn we have 1000(*t* + 1) + L = ___2__7 in the grid, so L ends in 7 and its cube root ends in 3; 3^{3} = 27 is already used and 23^{3} = 12167 is too big, so **L = 2197** = 13^{3} and **6dn = 3197**.
Now 9dn = 1000(*t* + 1) + 2197 must end in 2197 (using the 2 already in the grid), so *t* is ≥ (1102197 − 2197)/1000 − 1 = 1099 and ≤ (9992197 − 2197)/1000 − 1 = 9989, always ending in 9, and its cube root, also ending in 9, is between ∛1099 ≈ 10.3 and ∛9989 ≈ 21.5.
Thus the cube root can only be 19, making ** t = 6859** and

For 22ac we now have *y* − 750 to fit 9_9_ in the grid, so the cube root of *y* is between ∛(9090 + 750) ≈ 21.4 and ∛(9999 + 750) ≈ 22.1, namely 22, making ** y = 10648** and

For 20ac, S − 125 has three digits, so S’s cube root is between ∛(110 + 125) ≈ 6.2 and ∛(999 + 125) ≈ 10.4, ie one of {7, 8, 9} (not 10, because the cube 1000 is already used).
We can rule out 9 because 729 + 125 = 604 would make 21dn start with 0, so S is one of {343, 512}.
From 12dn, 9F + S = 3___8_, so the cube root of F is between ∛((301081 − 512)/9) ≈ 32.2 and ∛((399989 − 343)/9) ≈ 35.4, ie one of {33, 34, 35}, with F in {35937, 39304, 42875} respectively.
Considering the last two digits only, 9F ends in one of {33, 36, 75} and S ends in {12, 43}; the only combination of those that can give a sum of 8_ for the end of 12dn is 75 + 12, corresponding to **F = 42875**, **S = 512**, **12dn = 386387** and **20ac = 387**.

For 25ac we have D − 539 = 78___ in the grid, so the cube root of D is between ∛(78000 + 539) ≈ 42.8 and ∛(78999 + 539) ≈ 43.01, ie it’s 43, **D = 79507** and **25ac = 78968**.
Now 11ac = 79480 − 2*o* = 87_ in the grid, so the cube root of *o* is between ∛((79480 − 879)/2) ≈ 33.999 and ∛((79480 − 870)/2) ≈ 34.0003, ie it’s 34, ** o = 39304** and

For 18dn we have B − 872 = _3_9 in the grid, so B and its cube root end in 1; 11^{3} − 872 = 459 is too small and 31^{3} − 872 = 28919 is too big, so **B = 9261** = 21^{3} and **18dn = 8389**.
Now we have 13dn = *e* + 13357 = _8_82 in the grid, so *e* and its cube root end in 5; the cube root is between ∛(18182 − 13357) ≈ 16.9 and ∛(98982 − 13357) ≈ 44.1, ie one of {25, 35}, but 35^{3} = 42875 is already taken by F, so ** e = 15625** = 25

For 10ac we have 2T + 1512 = _198 in the grid, so T ends in 3 and its cube root ends in 7.
The root is < ∛((9198 − 1512)/2) ≈ 15.7, so it must be 7, with **T = 343** and **10ac = 2198**.
Now 21dn = 1072 − *s* for 8_6 in the grid, so *s* and its root end in 6.
The root is < ∛(1072 − 806) ≈ 6.4, ie it’s 6, with ** s = 216** and

Now 13ac = 6264 − N = 2_8_ in the grid, so the cube root of N is between ∛(6264 − 2989) ≈ 14.9 and ∛(6264 − 2080) ≈ 16.1, ie one of {15, 16}.
The corresponding values for 5dn are {15^{3}×216 − 64 = 728936, 16^{3}×216 − 64 = 884672}, but the grid has _28___, so **N = 3375** = 15^{3}, **13ac = 2889** and **5dn = 728936**.
That completes **17ac = 6383** in the grid and from its clue 9045 − 2*l* we can get ** l = 1331**.
It also completes

For 9ac we have I + 3375 = 61_9, so I and its root end in 4 and I ≤ 6199 − 3375 = 2824; 4^{3} = 64 is already used and 24^{3} = 13824 is too big, so **I = 2744** = 14^{3} and **9ac = 6119**.
For 1ac = W + 11334 we have _1_87 in the grid, so W ends in 53 and its root ends in 7.
The root is < ∛(91987 − 11334) ≈ 43.2, ie one of {17^{3} = 4913, 27^{3} = 19683, 37^{3} = 50653} (7^{3} = 343 is already taken), of which the only one ending in 53 is **W = 50653**, making **1ac = 61987**.

The grid is now full, as shown here, but we still don’t have the cubes for A, G, O and R.
In order of their values, the letters we have so far give {*b* Y J U *s* T S P E *l* L I N *i t* B *y* H *e d o* F W D}, from which “*b*Y JU*s*T SPE*l*LIN(G) *it*, B*y* H*e*(A)*d* *o*F W(OR)D” would be a reasonable guess, and enough to finish the puzzle.
But if we want to confirm the values, we have to test the remaining cubes, which are {1728, 4096, 4913, 8000, 13824, 17576, 19683, 21952, 24389, 27000, 29791, 35937, 46656, 54872, 59319, 64000, 68921, 74088, 85184, 91125, 97336}.
From 3dn we know R = O + 9128, and if we add 9128 to each of the remaining cubes the only cube sum we get is 54872 + 9128 = 64000, so **O = 54872** and **R = 64000**.
Similarly, 23ac tells us A = 2G + 9857, and plugging the remaining cubes into that gives only **G = 4913** and **A = 19683**.

The message is now confirmed as “*b*Y JU*s*T SPE*l*LING *it*, B*y* H*e*A*d* *o*F WORD”.
For the first part we need to change each grid digit by spelling it out, ie 1 = one, 2 = two etc.
Then we need to replace each of those words by its “head” (first letter) and the result is a grid full of real words.