the product of three consecutive integers is 22 less than the cube of the middle integer. Find the integer.

let the 3 consecutive integers be

x-1, x, and x+1

( I wanted to cube x , not x+1 )

x(x-1)(x+1) = x^3 - 22
x^3 - x = x^3 - 22
x = 22

the 3 integers are 21 , 22, and 23

check: product of 3 numbers = 21x22x23 = 10626
cube of 22 = 10648
is the product 22 less than the cube? YES

Let's assume the three consecutive integers as x, (x+1), and (x+2).

According to the given information, the product of the three consecutive integers is 22 less than the cube of the middle integer. We can represent this mathematically as:
x(x+1)(x+2) = (x+1)^3 - 22

Now, let's simplify the equation and solve for x:

x(x+1)(x+2) = (x+1)(x+1)(x+1) - 22

Expanding the right side of the equation:
x(x+1)(x+2) = (x+1)^3 - 22
x(x+1)(x+2) = (x+1)(x^2+2x+1) - 22
x(x+1)(x+2) = x^3 + 3x^2 + 3x + 1 - 22
x(x+1)(x+2) = x^3 + 3x^2 + 3x - 21

Now, let's simplify further:

x^3 + 3x^2 + 3x - 21 = x^3 + 3x^2 + 3x

Subtracting x^3 + 3x^2 + 3x from both sides of the equation:
-21 = 0

This equation does not have a solution. Therefore, there are no three consecutive integers where the product of the three integers will be 22 less than the cube of the middle integer.

To solve this problem, let's break it down step by step:

1. Let's assume the three consecutive integers as n-1, n, and n+1.

2. According to the given statement, the product of these three consecutive integers is 22 less than the cube of the middle integer. Mathematically, we can represent this as:
(n-1) * n * (n+1) = n^3 - 22

3. Expand the left side of the equation:
n^3 - n = n^3 - 22

4. Simplify the equation by canceling out the n^3 term on both sides:
- n = -22

5. Solving for n, multiply both sides by -1:
n = 22

So, the integer is 22.