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So all even \(n\) satisfy \(n^3 \equiv 0 \pmod{8}\). So condition is \(n \equiv 0 \pmod{2}\), but we’ll refine later.Step 2: Solve modulo 125n^3 \equiv 888 \pmod{125}Note \(888 \mod 125 = 888 - 7\cdot125 = 888 - 875 = 13\), so:n^3 \equiv 13 \pmod{125}We search for a solution to \(n^3 \equiv 13 \pmod{125}\). Try lifting via Hensel’s Lemma or trial.Try small values modulo 5 first:Mod 5: \(n^3 \equiv 13 \equiv 3 \pmod{5}\)Check cubes mod 5:\(0^3 = 0\)