I am not sure how to solve this.  The answer in the back of the book says 8,9,10 or -1,0,1

I was hoping someone can show me how to do the problem
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Let the integers be n-1, n and n+1. Their sum is 3n. The middle number squared is n^2. So 3n=n^2/3. Divide both sides by n (provided n not zero): 3=n/3 so n=9 and the integers are 8, 9 and 10. However, when we divided by n we assumed that n wasn't zero. Clearly if n=0 3n=n^2/3=0, and the equation is satisfied, so the three integers can also be -1, 0 and 1.

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