Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$

Question

I am a student and this question is part of my homework.

May you tell me if my proof is correct?

Thanks for your help!

Prove that $(z^3-z)(z+2)$ is divisible by $12$ for all integers $z$.

$(z^3-z)(z+2)=z(z^2-1)(z+2)=z(z-1)(z+1)(z+2)=(z-1)(z)(z+1)(z+2)$

$(z-1)(z)(z+1)(z+2)$ means the product of $4$ consecutive numbers.

Any set of $4$ consecutive numbers has $2$ even numbers, then $(z-1)(z)(z+1)(z+2)$ is divisible by $4$.

Any set of $4$ consecutive number has at least one number that is multiple of $3$, then $(z-1)(z)(z+1)(z+2)$ is divisible by 3.

Therefore $(z-1)(z)(z+1)(z+2)$ is divisible by $12$. Q.E.D.

Answer 1

That's correct. Alternatively it is divisible by $\,24\,$ by integrality of binomial coefficients

$$\,(z+2)(z+1)z(z-1)\, =\, 4!\ \dfrac{(z+2)(z+1)z(z-1)}{4!}\, =\, 24{ {z+2\choose 4}}\qquad\qquad$$

Similarly $\,n!\,$ divides the product of $\,n\,$ consecutive naturals.