Sum of cubes of two rational but not of two integers

Question

Is there an integer which is the sum of the cubes of two rational numbers, but not the sum of the cubes of two integers?

This is not possible if we consider cubes in place of squares (Davenport-Cassels theorem, see here, here), but what about my situation? Notice that $7$, see here, is not the sum of the cubes of two rational numbers, but every rational number is the sum of the cubes of three rational numbers.

Answer 1

If $n$ is any integer then $n^3 \equiv -1,0,1 \pmod 9$. Hence if $a,b$ are any integers:

$$a^3 + b^3 \equiv -2,-1,0,1,2 \pmod 9$$ Conversely if any integer $k \equiv 3,4,5,6 \pmod 9$, it cannot be a sum of two cubes of integers. Searching among small integers meeting this condition, I found:

$$\Big(\frac{2}{3}\Big)^3 + \Big(\frac{7}{3}\Big)^3 = 13$$

Answer 2

$$(\frac{17}{21})^3+(\frac{37}{21})^3=6$$ which is neither the sum nor the difference ot two integer-cubes.