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

check:

product of the three:

3(4)(5) = 60

cube of the smallest = 27

difference = 60-27 = 33

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