Proof without words: The difference of consecutive cubes

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For a more conventional algebraic proof, notice that

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

The product n(n+1) is always an even number times an odd number: if n is even, then n+1 is odd, but if n is odd, then n(n+1). So n(n+1) is a multiple of 2, and so 3n(n+1) is a multiple of 6. Therefore, 3n(n+1)+1 is one more than a multiple of 6, proving the theorem.

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