In the paper Characterizing the Sum of Two Cubes, Kevin Broughan gives the relevant theorem,
Theorem: Let $N$ be a positive integer. Then the equation $N = x^3 + y^3$ has a solution in positive integers $x,y$ if and only if the following conditions are satisfied:
There exists a divisor $m|N$ with $N^{1/3}\leq m \leq (4N)^{1/3}.$
And $\sqrt{m^2-4\frac{m^2-N/m}{3}}$ is an integer.
The sequence of integers $F(n)$,
$$\begin{aligned}
F(n)
&= a^3+b^3 = (2 n + 6 n^2 + 6 n^3 + n^4)^3 + (n + 3 n^2 + 3 n^3 + 2 n^4)^3\\
&= c^3+d^3 = (1 + 4 n + 6 n^2 + 5 n^3 + 2 n^4)^3 + (-1 - 4 n - 6 n^2 - 2 n^3 + n^4)^3
\end{aligned}$$
for integer $n>3$ apparently is expressible as a sum of two positive integer cubes in exactly and only two ways.
$$\begin{aligned}
F(4) &= 744^3+756^3 = 945^3+15^3\\
F(5) &= 1535^3+1705^3 = 2046^3+204^3\\
&\;\vdots\\
F(60) &=14277720^3+26578860^3 = 27021841^3+12506159^3
\end{aligned}$$
Using Broughan's theorem, I have tested $F(n)$ from $n=4-60$ and, per $n$, it has only two solutions $m$, implying in that range it is a sum of two cubes in only two ways. Can somebody with a faster computer and better code test it for a higher range and see when (if ever) the proposed statement breaks down? Incidentally, we have the nice relations,
$$a+b = 3n(n+1)^3$$
$$c+d = 3n^3(n+1)$$
Note: $F(60)$ is already much beyond the range of taxicab $T_3$ which is the smallest number that is the sum of two positive integer cubes in three ways.
$$T_3 \approx 444.01^3 = 167^3+436^3 = 228^3+423^3 = 255^3+414^3$$
(Using the theorem, this yields 3 values for $m$.)
It is a standard conjecture in Number Theory that there exist infinitely many $s$ such that $p=10s+1$, $q=15s+2$, and $r=6s+1$ are all prime. Then $3p=30s+3$, $2q=30s+4$, and $5r=30s+5$ are consecutive integers, each a product of two primes, so $\mu(3p)=\mu(2q)=\mu(5r)=1$.
Best Answer
Any number $m\equiv\pm4\pmod9$ cannot be expressed as a sum of $3$ cubes:
No $3$ values chosen from $\{0,+1,-1\}$ will ever sum up to $\pm4\pmod9$.