Problem
NT-B1-M11-P006 Cube and Number
#6
★★☆☆☆ Level 2 of 5
Prove that \(3\mid n^3-n\) for every integer \(n\).
Factor \(n^3-n\).
We have \(n^3-n=n(n-1)(n+1)\). Among three consecutive integers, one is divisible by \(3\), so the product is divisible by \(3\).
Standard task on choosing factorisation.