Just mod 7 and bounding

Take $\mod{7}$ gives $3|n$, now let $n=3k$ we have \[(11^k)^3<11^{3k}+8^k+6<(11^k+1)^3=11^{3k}+3\times 11^{2k}+3\times 11^k+1\]which means $11^{3k}+2^{3k}+6$ is between two cubes, so there is no solution.

Hamzaachak wrote: Just mod 7 and bounding What's your motivation of using mod 7 instead of any other modulos

m4thbl3nd3r wrote: Hamzaachak wrote: Just mod 7 and bounding What's your motivation of using mod 7 instead of any other modulos The cubic residues of $7$ are just $0,1,6$. While for other modulos the cubic residue set might the same as complete residue set (for example, mod $11$)