I came along the problem of finding three perfect cubes that are consecutive numbers of an arithmetic progression, i.e: $a^3-b^3=b^3-c^3$, where $a>b>c$ (to avoid trivial solutions). Clearly it is equivalent to solve $x^3+y^3=2$ over the rationals.
This is something I tried:
Consider the curve $x^3+y^3=2$ in te plane. I looked for a rational parametrisation of it. Ok, maybe looking for a rational parametrisation won't give all solutions (I'm not sure about that) but it's a start and we'll see what it leads us to.
I tried considering the intersection points with the line $y=tx+(1-t)$, since we now $(1,1)$ is such a point. (And I had seen a similar approach to find solutions to $x^2+y^2=1$.)
Solving the system for $x$ gave me (after factoring out the superfluous factor $x-1$) a quadratic in $x$ with discriminant $-3(t+1)^4+36t^2-4$. So it suffices to figure out when this is the square of a rational.
Let's write it as $-3(p+q)^4+36p^2q^2-4q^2=d^2$, where $p=qt$.
And here's where I can't continue. Any proceeding of this method or other methods are welcome.
P.S: This is not homework, it's just a question out of curiosity.