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Could someone help me with my graded math homework?

The question is,
The product of three consecutive integers is 21 more than the cube of the smallest integer. Find the integers.
if you can help me that's be fantastic! and if you could, please step out the work on here? THANK YOU SO MUCH!

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I would write it out like this, eerything in terms of A. The first integer would be A and because they are consecutive integers, the other integers would be A+1 and A+2. All of them multiplyed together = the cube of A + 21

A(A+1)(A+2)=A^3+21

string it all out...

(A^2+A)(A+2)=cbrt A +21 A^3 + 2A^2 + A^2 + 2A= A^3 +21

Subtract A^3 from each side and combine like terms:

3A^2+ 2A=21

Subtract 21 from both sides to equal everything to zero

3A^2 + 2A -21 = 0

need to know your factoring skills here... factor the expression:

(3A-7)(A+3)=0

solve for A: A = 7/3 and -3

So your three consecutive integers would be -3, -2, -1 (7/3 is not an integer)

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But I'm confused- the product of -3, -2, -1 is -6. -3^3 is -27. 21 more than -6 is +15. Isn't it?
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you said the product of the three consecutive numbers needs to be 21 more than the cube of the smallest integer.

well the product of the numbers i gave you is -6
the cube of the smallest number (-3) is -27

-6 is 21 more than -27
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I realized today that you were right haha it was late last night but thank you! :)
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I messed up. Haha

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19,2, and 0 aren't consecutive integers but thanks for trying!
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Consecutive integers = integers with a difference of 1. Sorry, I wasn't thinking
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Haha it's fine!
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Let n = 1st integer
n+1 = 2nd
n+2 = 3rd
n(n + 1)(n + 2) = n^3 + 21
n(n^2 + 3n + 2) = n^3 + 21
n^3 + 3n^2 + 2n = n^3 + 21
3n^2 + 2n - 21 = 0
(3n - 7)(n + 3) = 0
3n-7=0 or n+3=0
n=7/3. n= -3
Throw out n=7/3 because it's not an integer.
So n= -3, n+1= -2, n+2= -1. ANSWER

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