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Consecutive Cube

Problem

Three consecutive integers are multiplied together and the middle number is added.
E.g. 3 times 4 times 5 = 60 and 60 + 4 = 64 = 43

Will this always produce a cube number?


Solution

Taking three consecutive integers, n, n + 1 and n + 2 we get.

n(n + 1)(n + 2) + (n + 1)
= (n + 1)(n(n + 2) + 1)
 = (n + 1)(n2 + 2n + 1)
 = (n + 1)(n + 1)2
 = (n + 1)3

Problem ID: 83 (Oct 2002)     Difficulty: 2 Star

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