find three consecutive integers whose product is 33 larger than the cube of the smallest integer
if the three numbers are x-1,x,x+1, then
(x-1)(x)(x+1) = x^3+33
x^3-x = x^3+33
x = -33
(-34)(-33)(-32) = -35904 = -33^3 + 33
To find three consecutive integers, let's start by assuming the first integer as "x".
The second integer will be x + 1 and the third integer will be x + 2.
According to the given condition, their product is 33 larger than the cube of the smallest integer.
So, we can write the equation as:
x * (x + 1) * (x + 2) = x^3 + 33
Now, let's solve the equation step by step:
Expand the left side of the equation:
x^3 + 3x^2 + 2x = x^3 + 33
Simplify by subtracting x^3 from both sides:
3x^2 + 2x = 33
Subtract 33 from both sides:
3x^2 + 2x - 33 = 0
Now, we need to solve this quadratic equation to find the value of x.
This equation can be factored as:
(3x - 11)(x + 3) = 0
Setting each factor equal to zero:
3x - 11 = 0 or x + 3 = 0
By solving these equations, we get:
x = 11/3 or x = -3
Since we are looking for three consecutive integers, we will ignore the negative value for x. Therefore, we take x = 11/3.
Now, we have the smallest integer (x) = 11/3.
The three consecutive integers are:
11/3, 11/3 + 1, and 11/3 + 2
To get the exact values:
11/3 = 3.67
11/3 + 1 = 4.67
11/3 + 2 = 5.67
So, the three consecutive integers whose product is 33 larger than the cube of the smallest integer are approximately 3.67, 4.67, and 5.67.