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Question 718010: The product of three consecutive integers is 21 more than the cube of the smallest integer. Find the integers.
Found 2 solutions by DrBeeee, algebrahouse.com:
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Let n = first integer
Then the second and third consecutive integers are n+1 and n+2, respectively.
Your problem statement gives us
(1) n*(n+1)*(n+2) = n^3 - 21, agree?
Simplify (1) to get
(2) n*(n^2 +3n +2) = n^3 + 21 or
(3) n^3 + 3n^2 + 2n = n^3 + 21
The n^3 terms of (3) cancel leaving
(4) 3n^2 + 2n - 21 = 0
Are you good at factoring? No, use the quadratic factoring equation.
I'm fair enough at factoring to be able to use the basic "rules" and get
(5) 3n^2 + 2n -21 = (3n - x)*(n + y), where x*y = 21, try 7*3 and get
(6) 3n^2 + 2n -21 = (3n-7)*(n+3)
If you FOIL the right side of (6) you wili get the left side of (6).
Setting (6) equal to zero you get the two roots of (4) as
(7) n = (7/3,-3).
The first root, 7/3 is not a solution because 7/3 is not an integer. The correct root is
(8) n = -3
Let's check the value n = -3 with (1).
Is (-3*-2*-1 = (-3)^3 + 21)?
Is (-6 = -27 +21)?
Is (-6 = -6)? Yes
Answer: The three consecutive integers are -3, -2, and -1.
Answer by algebrahouse.com(1659)  (Show Source):
You can put this solution on YOUR website! x = smallest integer
x + 1 = second integer
x + 2 = third integer {consecutive integers increase by 1}
x(x + 1)(x + 2) = x³ + 21 {product of the three is 21 more than cube of smallest}
(x² + x)(x + 2) = x³ + 21 {used distributive property}
x³ + 3x² + 2x = x³ + 21 {used foil method}
3x² + 2x = 21 {subtracted x³ from each side}
3x² + 2x - 21 = 0 {subtracted 21 from each side}
3x² + 9x - 7x - 21 = 0 {split 2x into 9x and -7x}
3x(x + 3) - 7(x + 3) = 0 {factored 3x out of first two terms and -7 out of last two terms}
(3x - 7)(x + 3) = 0 {factored x + 3 out of the two terms}
3x - 7 = 0 or x + 3 = 0 {set each factor equal to 0}
x = 7/3 or x = -3 {solved each equation for x}
x = -3 {fractions are not integers}
x + 1 = -2 {substituted -3, in for x, into x + 1}
x + 2 = -1 {substituted -1, in for x, into x + 2}
-3, -2, and -1 are the three consecutive integers
For more help from me, visit: www.algebrahouse.com
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