Alzebra by Rekha Kushwaha

If $d\neq 0,a$ are integers, then $\frac{(a+nd)^3 - a^3}{d} = \frac{3a^2nd + 3an^2d^2 + n^3d^3}{d} = 3a^2n + 3adn^2 + d^2n^3$ is a positive integer for infinitely many integers $n$ (since it's cubic in $n$ )

Now, if we suppose $a^3$ is a cube that is in an arithmetic progression with common difference $d\neq 0$ (which needs to at least be rational, but we may assume is an integer), then the above shows that $(a+nd)^3 = a^3 + \big(3a^2n + 3adn^2 + d^2n^3\big)d$ is also in the arithmetic progression for any $n$ such that $3a^2n + 3adn^2 + d^2n^3 > 0$ , of which there are infinitely many.

Remarks:

• If $d=0$ then the arithmetic progression is constant, so every element is a cube as long as one is (of course, it's all the same cube, so there's a small issue of semantics, which is why I ignored that possibility.
• If $d = \frac{b}{c}$ is rational, then we can just consider the A.P. subsequence with common difference $b$
• If $d$ is irrational, then the sequence contains at most one integer, which could provide a problem, which is why I ignored it.