SOLUTION: find three consecutive integers whose product is 56 larger than the cube of the smallest integer

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Question 896003: find three consecutive integers whose product is 56 larger than the cube of the smallest integer
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
consecutive integers: (x-1), x, (x+1)

(x-1)(x)(x+1) = 56 + (x-1)^3
x^3 - x = x^3 - 3x^2 + 3x + 55
0 = -3x^2 + 4x + 55
0 = 3x^2 - 4x - 55
0 = (x - 5)(3x + 11)
x = 5 or x = -11/3

x has to be an integer, so the integers are 4, 5, and 6.