Find three consecutive intergers whose product is 33 larger than the cube of the smallest integer.

asked by Josh
  1. let the smallest be x
    the middle one x+1
    the largest x+2

    x(x+1)(x+2) - x^3 = 33
    x(x^2 + 3x + 2) - x^3 = 33
    x^3 + 3x^2 + 2x - x^3 - 33 = 0
    3x^2 + 2x-33=0
    (x-3)(3x + 11) = 0
    x = 3 or x = -11/3, but x must be an integer,
    so x = 3

    the smallest is 3 , the middle one is 4 and the largest is 5

    product of the three:
    3(4)(5) = 60
    cube of the smallest = 27

    difference = 60-27 = 33

    posted by Reiny

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