Seeking a combinatorial proof $\sum\limits _{k=0}^n (n-2k)^3\binom{n}{k}=0$

Question

I would appreciate if somebody could help me with the following problem

Q: Seeking a combinatorial proof $(\binom{n}{k}=\frac{n!}{k! (n-k)!} )$

$$\sum _{k=0}^n (n-2 k)^3 \binom{n}{k}=0$$

Answer 1

$$ 2\sum _{k=0}^n (n-2 k)^3 \binom{n}{k}= \sum _{p+q=n} \left[ (n-2 p)^3 \binom{n}{p}+(n-2 q)^3 \binom{n}{q} \right] \\ =\sum _{p+q=n} \left[(n-2 p)^3 \binom{n}{p}+(2p-n)^3 \binom{n}{p} \right] \\ =0 $$