The values of the letters in clues can be determined using the lengths of the grid entries and the fact that none can start with zero, before any of the clue values are actually fitted into the grid. The message given by the numbers inserted into grid entries is “all threes eights and nines highlighted in green”; so all the cells containing any of {3, 8, 9} are shaded green, forming the word GO in big letters, which is hinted at by the puzzle’s title.

This is one possible solving path.

The entry at 1ac has six digits, so the answer must have four or five.
For A^{3} + A^{2} to be ≥ 1000, A must be 10 or 11.
If A is 10, the clue answer would be 1100,
but that would make at least one crossing down entry start with zero,
so **A = 11**
and the entry is 1452 with either 10 or 11 inserted.

14ac and 30dn are both three-digit entries,
so the clue answers E^{3} + H^{3} + S^{3}
and H^{3} + N^{3} + S^{3} are both < 100,
which means none of {E, H, N, S} can be ≥ 5
(because 5^{3} = 125),
and they are therefore {1, 2, 3, 4} in some order.

The answer to 27dn,
H^{3} + I^{3} − S^{3},
is < 100 (because the entry has three digits).
If I is 6 then the minimum possible value is
1^{3} + 6^{3} − 4^{3} = 153,
too big, so **I = 5**.
If S < 3, the minimum possible is
1^{3} + 125 − 2^{3} = 118, too big;
if S is 4, 25dn would be 4^{2} + 4 = 20,
making one of the across entries start with zero;
so **S = 3**.
Now 27dn is
H^{3} + 125 − 27 = H^{3} + 98 < 100,
which forces **H = 1**
and leaves {E, N} = {2, 4} in either order.

The answer to 19dn is 5(5G^{2} − 11) < 1000,
so G < √((1000/5 + 11)/5) ≈ 6.5,
which makes **G = 6**.
For 3dn, L^{2}(11 + 5 + 3) − R < 1000.
If L > 7, the minimum possible is
8^{2}×19 − 10 = 1206,
too big; so **L = 7**.

For 9dn, 11E(3D − 1) < 1000.
If E is 4, the minimum possible is
11×4(3×8 − 1) = 1012, too big,
so **E = 2** and **N = 4**.
For 12dn,
(2×6×1×5×R − 3)(4 + 7 + R) < 10000.
If R > 8, the minimum is
(60×9 − 3)(11 + 9) = 10740, too big,
so **R = 8**.

That leaves {D, T} = {9, 10} in either order.
If T is 10, 1dn would be 4(5×7 + 10) = 180,
making either 15ac or 18ac start with zero.
So **T = 9** and **D = 10**.

Now that we know all the letter values, we can calculate all the clue answers and start fitting them into the grid. We'll use the notation [x] to denote the inserted digits in an entry.

1ac = 1452 with two digits inserted.
If they are [10],
the 0 must be in the unchecked third cell
(to avoid a down entry starting with zero),
making 2dn start with 1;
but the answer to 2dn is 28,
and if it were entered as [1]28
the first cell would contain an inserted digit in both entries,
which is disallowed.
Therefore 1ac has [11] inserted somewhere;
it can't be [11]1452 or 1[11]452,
because 2dn can't start with an inserted [1], as above;
similarly, it can't be 14[11]52 or 145[11]2,
because 3dn = 923 can't start with an inserted [1];
therefore **1ac = 1452[11]**,
as the answers to both 4dn and 5dn start with 1.
That forces **2dn = [4]28**
and **3dn = [2]923**.

15ac = 5622 now has 8 in the second cell,
so **15ac = 5[8]622**,
which forces **1dn = 17[5]6**
and **12dn = 9[6]063**,
completing **11ac = [7]299**.
Now 18ac can only be entered as **18ac = 690[3]**,
which forces **19dn = [9]845**.
That in turn forces **25ac = 18[6]2**
and **31ac = 64[3]**,
and then **25dn = 1[6]2**.

26dn = 20
and the entry's middle digit can't be zero (at the start of 32ac),
so the entry is **26dn = 2[8]0**,
inserting the first digit of 32ac.
34ac is 2522 with two digits inserted,
and the entry has 0 for the fourth digit,
so **34ac = 25[10]22**.
27dn is 99 but the entry ends in 2,
so **27dn = 99[2]**.
Similarly, 21dn is 811 ending in 2,
so 21dn = **811[2]**.
Then 27ac is 95, to fit 91_,
so **27ac = 9[1]5**.

The first digit of the entry at 22dn has to be
the (correct) last digit of 20ac (988)
or the first digit of 22dn (83);
either way, it's 8.
Then the entry's second digit must be
the 3 from 22dn or the 3 from the start of 29ac = 3512,
ie it's 3, so 22dn is 83[1_].
The inserted last digit must match
the correct first or second digit of 35ac (20740),
which forces **22dn = 83[10]**.
The entry for 32ac = 891 is now 891_1,
with the last digit being an inserted one in 22dn,
so it must be 89[1_]1.
The missing digit must be a real digit of 28dn (51),
so **32ac = 89[11]1**
and **28dn = 51[2]**,
inserting the 2 from the start of 35ac.

If 24dn = 240 is entered with the 2 in the first cell,
then 23ac is 184[2],
which forces 14dn to be 32[8]59.
That in turn makes 29ac = 35[5]12 (not 3[5]512),
making 23dn = 1[5]34 and 24dn = 2[2]40,
which forces 33ac to be 392[4].
But then 30dn is 12_, which can't fit 92;
therefore the first cell in 24dn contains the 4 from 23ac,
ie **24dn = [4]240**.

That then makes **33ac = 392[4]**,
and 30dn is _2_,
which can only be 9[2]2 or 92[4].
Now 29ac has 9 as the fourth digit,
so **29ac = 351[9]2**,
which forces **14dn = 325[1]9**
and **23dn = 1[5]34**,
and then **35ac = 20[4]740**
(completing **30dn = 92[4]**)
and **23ac = 1[5]84**.

17ac has 2 as the second digit,
so **17ac = 1[2]35**,
which forces **8dn = 19[1]**
and **10dn = 21[5]**,
which respectively lead to **13ac = 3755[9]**
and **14ac = 36[1]**.
The only possibility for 6dn is **6dn = 7[5]22**,
which forces **6ac = [7]61772**
(completing **7dn = [6]58**
and **9dn = [9]638**),
**16ac = 1[2]8**
(completing **5dn = 17[1]**)
and **20ac = 98[2]8**
(completing **4dn = 1[3]29**).

With the entries all deduced, their inserted numbers in clue order are {11, 7, 7, 9, 1, 8, 2, 2, 3, 2, 5, 6, 1, 9, 3, 11, 4, 10, 4, 5, 4, 2, 3, 1, 5, 6, 1, 7, 5, 6, 1, 9, 2, 10, 5, 4, 6, 8, 2, 2, 4}, which translate to ALL THREES EIGHTS AND NINES HIGHLIGHTED IN GREEN.