Question 981050
The sum of the products of three consecutive integers taken twice at a time is 107.  Find the integers.
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n-1, n, n+1 are the integers
n(n-1) + n(n+1) + (n-1)*(n+1) = 107
2n^2 + n^2 - 1 = 107
3n^2 = 108
n = 6
n = -6