[Spivak]

The easy way of seeing the first part is to do the prime factorization. The 7 doesn't matter since it's prime. If n has a 2 in its factorization it now has 2^3. But if it doesn't have a 2 it won't suddenly acquire one.

All the symbol soup proofs aren't wrong but I don't think they satisfyingly explain the why.

[thaumasiotes]

All of these "always divisible by n" proofs are asking you to solve them case by case in modular arithmetic.

For divisibility by two, there are only two cases. So if n is 1, then n³ is 1, and if n is 0, n³ is 0. 0+0 = 0; 1+1 = 0; and this completes the proof.

I am not actually sure that doing a prime factorization on 7n³ for unknown n is easier than knowing that 1³ = 1.