# Proof without words: The difference of consecutive cubes

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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