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

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> 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      Log On


   

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) About Me  (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:

x%5E3%2B3x%5E2%2B2x+=+x%5E3%2B21

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) About Me  (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