Combinatorial Proofs for $n^3-n= 6\binom{n}{2}+ 6\binom{n}{3}$

Question

I am learning about combinatorial proofs and I found it very interesting. One proof can be elegantly done while algebraic proofs can be tedious. However, I had a hard time come up with a combinatorial proof for the following equality. (My thought is that $n^3-n$ is really $P(n+1,3)$ and I can kind of see how 6 is coming (since $6 = 3!$) but then I don't know where to go next)

$n^3-n= 6C(n,2) + 6C(n,3)$

and I'd appreciate if you can shed some light so that I can keep going. Thanks ahead!

Answer 1

Divide through by $6$, and you can rewrite it as $$\binom{n+1}3=\binom{n}2+\binom{n}3\;,$$ which is just a special case of the Pascal’s triangle identity, which in turn has a straightforward combinatorial proof.