math
posted by Josh on .
Find three consecutive intergers whose product is 33 larger than the cube of the smallest integer.

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 + 2x33=0
(x3)(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 = 6027 = 33