Take two consecutive integers:

n, n + 1

The difference of their cubes is:

(n + 1)^3 - n^3

n^3 + 3n^2 + 3n + 1 - n^3

Cancel the n^3

3n^2 + 3n + 1

Factor out a 3 from the first 2 terms:

3(n^2 + n) + 1

The first two terms are always divisible by 3 but then the + 1 makes this expression not divisible by 3:

3(n^2 + n) + 1 = 1 (mod 3)

n, n + 1

The difference of their cubes is:

(n + 1)^3 - n^3

n^3 + 3n^2 + 3n + 1 - n^3

Cancel the n^3

3n^2 + 3n + 1

Factor out a 3 from the first 2 terms:

3(n^2 + n) + 1

The first two terms are always divisible by 3 but then the + 1 makes this expression not divisible by 3:

3(n^2 + n) + 1 = 1 (mod 3)

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