SOLUTION: The product of three consecutive integers is 21 more than the cube of the smallest integer. The smallest integer is A) -3 B) -4 C) -5 D) -6

Question 1132448: The product of three consecutive integers is 21 more than the cube of the smallest integer. The smallest integer is
A) -3
B) -4
C) -5
D) -6
Found 2 solutions by greenestamps, Alan3354:
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Let the three integers be x, x+1, and x+2. Then

the product of the three integers is x(x+1)(x+2) = x(x^2+3x+2) = x^3+3x^2+2x; and
the cube of the smallest integer is x^3.

The problem says the product of the 3 is 21 more than the cube of the smallest:

The x^3 terms cancel, leaving a quadratic equation which is easily solved by any of a number of different methods.

Of course, if an algebraic solution is not required, the answer is easily found by trying the given answer choices....
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The product of three consecutive integers is 21 more than the cube of the smallest integer. The smallest integer is
A) -3
B) -4
C) -5
D) -6
----------------
(x-1)*x*(x+1) = (x-1)^3 + 21
x^3 - x = x^3 - 3x^2 + 3x - 1 + 21
-x = -3x^2 + 3x + 20
3x^2 - 4x - 20 = 0
(3x - 10)*(x + 2) = 0
x = -2
--> -3 is the smallest integer

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