Three consecutive shifted cube remainders

Prove that for every positive integer n>8, with t=floor(n^{1/3}), the three integers n-t^3, n-(t-1)^3, and n-(t-2)^3 are not all cubes.

Background

Theorem 3.3 leaves unresolved the case in which two adjacent cube remainders are cubes. The authors explain that ruling out three consecutive shifted cube remainders that are all cubes would substantially strengthen their cube bootstrapping result.

References

Let $n>8$ be a positive integer, and let $t=\lfloor n{1/3}\rfloor$. Then $n-t3$, $n-(t-1)3$, and $n-(t-2)3$ are not all cubes.

— Computational results on sums of a prime with squares or cubes  (2609.20505 - Applegate et al., 17 Sep 2026) in Section 6, first conjecture

For any positive integers $a, b, c, t$, the integers $t3+a3$, $(t-1)3+b3$, and $(t-2)3+c3$ are not all equal.

— Computational results on sums of a prime with squares or cubes  (2609.20505 - Applegate et al., 17 Sep 2026) in Conjecture 6.2, Section 6