Question 1063664
What are three consecutive integers whose product is 693 more than their sum?
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{{{root(3,693)}}} =~ 9, the middle integer
--> 8, 9 & 10
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If you do it the hard way:
(n-1)*n*(n+1) = n-1 + n + n+1 + 693
n^3 - n = 3n + 693
n^3 - 4n - 693 = 0
n = 9