Sums of three cubes of form $a^3+b^3+c^3=(c+1)^3$, Part 2.

Let $(a,b,c,d)$ be positive. In a previous question, we asked for parameterizations to,

$$a^3+b^3+c^3 = (c+1)^3=d^3$$

where $c$ is a polynomial of deg-$n$ for $n>3$.

Question: We've found $c$ of various deg-$n$, but now we give it sharper focus: Is there of deg-4 or deg-8?

I. Degree 3

There are two well-known solutions where $c$ is of deg-3, namely,

$$(n)^3 + (3n^2 + 2n + 1)^3 + (3n^3 + 3n^2 + 2n)^3 = (3n^3 + 3n^2 + 2n + 1)^3$$ $$(3 n^2)^3 + (6 n^2 - 3n + 1)^3 + (9 n^3 - 6n^2 + 3n - 1)^3 = (9 n^3 - 6n^2 + 3n)^3$$

with $n=1$ yielding,

$$1^3 + 6^3 + 8^3 = 9^3$$ $$3^3 + 4^3 + 5^3 = 6^3$$

But with some tweaking, we can make $(3,4,5,6)$ appear explicitly as the constant terms in the second identity,

$$(3n^2 + 6n + \color{red}3)^3 + (6n^2 + 9n + \color{red}4)^3 + (9n^3 + 21n^2 + 18n + \color{red}5)^3 = (9n^3 + 21n^2 + 18n + \color{red}6)^3$$

II. Degree 6

Adam Bailey found one where $(3,4,5,6)$ appears as the constant terms again,

$$(27n^4+54n^3+45n^2+19n+\color{red}3)^3+(54n^4+81n^3+63n^2+23n+\color{red}4)^3+(243n^6+567n^5+675n^4+468n^3+201n^2+49n+\color{red}5)^3+(243n^6+567n^5+675n^4+468n^3+201n^2+49n+\color{red}6)^3$$

And Old Peter found the special case which share the same sum,

$$(9n^3)^3+(27n^4+18n^3+9n^2+3n+1)^3+c^3 = (c+1)^3\\ (9n^3+9n^2+3n+1)^3+ (27n^4+18n^3+9n^2+3n)^3+ c^3 = (c+1)^3$$

with $c=81n^6+81n^5+54n^4+27n^3+9n^2+3n.$ In fact, there are infinitely many deg-6 parameterizations to,

$$A^3+B^3+c^3 = a^3+b^3+c^3 = (c+1)^3$$

as shown in this answer. In the next section, turns out there's also a deg-12 solution to that special case.

III. Higher Degrees

There are actually $c$ with either deg-10 or deg-12. I didn't realized it then, but in one of my posts, Jan Magnus-Okland answered and gave 6 solutions (unformatted for easier copy-paste)

a = 6 n + 18 n^2 + 180 n^3 + 540 n^4 + 3240 n^5 + 5832 n^6 + 34992 n^7, 
b = 1 + 36 n^2 - 72 n^3 + 864 n^4 - 3240 n^5 + 5832 n^6 - 34992 n^7, 
c = 36 (n^2 + 42 n^4 + 1116 n^6 + 16524 n^8 + 104976 n^10)

a = 6 n^2 - 36 n^3 + 36 n^4 + 648 n^6 - 1296 n^7, 
b = 1 - 6 n^2 - 36 n^3 + 180 n^4 - 2592 n^7 + 3888 n^8, 
c = 6 (-n^2 - 6 n^3 + 30 n^4 - 864 n^7 + 1944 n^8 + 3888 n^10 - 23328 n^11 + 23328 n^12)

a = 6 n^2 + 36 n^3 + 36 n^4 + 2592 n^7 + 3888 n^8, 
b = 1 + 12 n^2 + 72 n^4 + 648 n^6 + 1296 n^7, 
c = 12 (n^2 + 6 n^4 + 108 n^6 + 216 n^7 + 324 n^8 + 1944 n^10 + 11664 n^11 + 11664 n^12)

a = -12 n^2 + 144 n^4 - 648 n^6 - 1296 n^7 + 7776 n^8, 
b = 1 - 6 n^2 + 36 n^3 + 180 n^4 - 648 n^6 + 1296 n^7 + 7776 n^8, 
c = 6 (-n^2 + 6 n^3 + 30 n^4 - 216 n^6 + 432 n^7 + 3240 n^8 - 7776 n^10 + 93312 n^12)

a = 6 n + 18 n^2 + 252 n^3 + 756 n^4 + 6480 n^5 + 23328 n^6 + 69984 n^7 + 314928 n^8, 
b = 1 + 6 n + 54 n^2 + 180 n^3 + 2268 n^4 + 3240 n^5 + 40824 n^6 + 34992 n^7 + 629856 n^8, 
c =  6 (n + 9 n^2 + 54 n^3 + 522 n^4 + 2052 n^5 + 16740 n^6 + 48600 n^7 + 396576 n^8 + 734832 n^9 + 5353776 n^10 + 5668704 n^11 + 51018336 n^12)

a = 36 n^2 + 72 n^3 + 1944 n^4 + 3240 n^5 + 40824 n^6 + 34992 n^7 + 629856 n^8, 
b = 1 + 6 n + 54 n^2 + 252 n^3 + 2052 n^4 + 6480 n^5 + 23328 n^6 + 69984 n^7 + 314928 n^8, 
c = 6 (n + 9 n^2 + 54 n^3 + 522 n^4 + 2052 n^5 + 16740 n^6 + 48600 n^7 + 396576 n^8 + 734832 n^9 + 5353776 n^10 + 5668704 n^11 + 51018336 n^12)

the last pair sharing the same $c$, so the special case $A^3+B^3+c^3 = a^3+b^3+c^3 = (c+1)^3$.

IV. Question

Q: To repeat the question in the first section: Are there parameterizations where $c$ is of deg-4 or deg-8? Seems odd there are for $(3,6,10,12)$ but skips those two even numbers. And it can't be that $n=4m$ is forbidden since there are $n=12$.

P.S. For clarity, we are not looking for deg-$n$ with $n > 12$. Small degrees first.