SOLUTION: Find three consecutive integers whose product is 208 larger that the cube of the smallest integer

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Question 902781: Find three consecutive integers whose product is 208 larger that 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) = 208 + (x-1)^3
x^3 - x = 208 + (x^3 - 3x^2 + 3x - 1)
0 = -3x^2 + 4x + 207

x is integer ---> x = 9

The integers are 8, 9, and 10.