Can del be cycled in a scalar triple product?

Question

If w is a vector field and x is the position vector, Is this allowed?

Answer 1

No. I will show you an example, where I will denote the position vector with $\vec{r}$ to avoid confusions:

Take $\vec{\omega}=[1, 0, 0]^T$, then $\vec{\omega}\times \vec{r}=[1, 0, 0]^T\times [x, y, z]^T=[0, -z, y]^T$. Now, calculate $\vec{\omega}\times(\vec{\omega}\times \vec{r})$: $[1, 0, 0]^T \times [0, -z, y]^T = [0, -y, -z]^T$.
Now take its divergence: $\partial_x0-\partial_yy-\partial_zz=-2$

Now let's work on your second expression:
$\text{rot}([1, 0, 0]^T)=[0, 0, 0]^T$ because it does not depend on $\vec{r}$. Now, you are dotting a vector and a zero vector, so it's result will be $0$.

$0 \neq -2$, so your identity is not true.