Question 813470
let x = the median integer; then the other two integers are (x-1), x+1)
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the difference between the cubes of the two larger of the three consecutive integers is 66 more than the difference between the cubes of two smaller integers.
(x+1)^3 - x^3 = x^3 - (x-1)^3 + 66
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Cube (x+1) and (x-1)
x^3+3x^2+3x+1 - x^3 = x^3 - (x^3-3x^2+3x-1)+ 66
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removing the brackets changes the signs
x^3 + 3x^2 + 3x + 1 - x^3 = x^3 - x^3 + 3x^2 - 3x + 1 + 66
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x^3's cancel (fortunately) 
3x^2 + 3x + 1  = 3x^2 - 3x + 67
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Combine like terms on the left, leave 66 where it is
3x^2 - 3x^2 + 3x + 3x + 1 - 1 = 66
6x = 66
x = 11 is the median integer
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:
See if that checks out
12^3 - 11^3 = 11^3 - 10^3 + 66
1728 - 1331 = 1331 - 1000 + 66
397 = 331 + 66