Problem
ALG-B1-M09-P002 Cube minus the number
#2
★☆☆☆☆ Level 1 of 5
Prove that \(n^3-n\) is divisible by \(6\) for every integer \(n\).
Write the expression as the product of three consecutive integers.
\(n^3-n=n(n-1)(n+1)\). Among three consecutive integers, one is divisible by \(3\), and one is divisible by \(2\). Therefore the product is divisible by \(6\).
A classical identity connecting algebra and divisibility.