Prove 24 divides $u^3-u$ for all odd natural numbers $u$

First note that $$ u^3-u = (u-1)u(u+1)$$

Given that $u$ is odd.In this case, $u-1$ and $u+1$ are even and one of them is divisible by 4. This follows from the basic observation that one of any two consecutive even numbers is divisible by 4. So, $(u-1)(u+1)$ is divisible by $4 \times 2 =8$.

Also, one of any three consecutive natural numbers is divisible by 3. So, one of $u-1,u,u+1$ is divisible by 3.

So, $(u^3-u)$ is divisible by 8 and 3, which are co-prime. So, it is divisible by $8\times 3=24$

Factor it: $u^3-u=u(u^2-1)=u(u-1)(u+1)$. Now show that one of the three factors must be divisible by $4$ and another by $2$, and that one must be divisible by $3$.

Observe that $u^3-u=(u-1)u(u+1)$. Now $3$ consequtive numbers are always divisible by $3$. So $3\mid u^3-u.$ Since $u$ is odd $\Rightarrow$ $u-1, \ u+1$ are even. Prove that one of $u-1, \ u+1$ is divisible by $4$ and the other by $2$. Conclude that $24\mid u^3-u.$