Posted by Josh on Wednesday, March 6, 2013 at 10:03pm.
Find three consecutive intergers whose product is 33 larger than the cube of the smallest integer.
- math - Reiny, Thursday, March 7, 2013 at 9:06am
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
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