How to arrange 1 to 15 such that the sum of any adjacent 3 numbers will be a perfect cube?

Question

The numbers 1 to 15 should be arranged in a way that any 3 adjacent numbers' sum will be a perfect cube.

Answer 1

It is impossible. Why?

Note that the largest possible sum of any three numbers from the set is $15+14+13=42<64=4^3$ while the smallest possible sum of any three numbers will be $1+2+3=6>1=1^3$, so the only two available cubes will be $8$ and $27$

Suppose we could arrange the numbers in such a way as to have all consecutive triples adding up to be a cube. Consider what happens to a number larger than $8$.

Let $a_1,a_2,a_3,a_4$ be distinct members of the set $\{1,2,\dots,15\}$ with specifically $a_2>8$.

Suppose $a_1,a_2,a_3,a_4$ satisfies that $a_1+a_2+a_3$ is a cube and that $a_2+a_3+a_4$ is a cube. Then $a_1+a_2+a_3>8$ and is a cube, so $a_1+a_2+a_3=27$ is the only possibility. Similarly $a_2+a_3+a_4=27$ is the only possibility, but then that implies that $a_1=a_4$, a contradiction.