How to check if a number can be represented as the sum of two consecutive perfect cubes.

If you have factors of $N$ then you may use the fact that $m^3+(m+1)^3=(m^2+m+1)(2m+1)$. Then the quadratic factor cannot have any $6k-1$ prime factors, so $2m+1$ must be a multiple of the product of all such factors appearing in $N$. For instance, if $N=35$ then $2m+1$ must be a divisor of $35$ that's a multiple of $5$, and if this $N$ is to be a sum of two consecutive cubes at all then $m$ can only be $2$ or $17$.