Compute the curl Given that $\vec{\nabla}.{\vec{m}}=0$, and the vectors be in $\mathbb{R}^3$ I am trying to show that $$\vec{\nabla}\times \frac{r^2}{2}\vec{r}\times\vec{m}=(\vec{m}.\vec{r})\vec{r}-2r^2\vec{m}$$
I did 
$$\frac{r^2}{2}\vec{\nabla}\times\vec{r}\times\vec{m}=\frac{r^2}{2}(\vec{r}(\vec{\nabla}.\vec{m})-\vec{m}(\vec{\nabla}.\vec{r})-(\vec{r}.\vec{\nabla})\vec{m}+(\vec{m}.\vec{\nabla})\vec{r})$$
$$=\frac{r^2}{2}(\vec{m}.\vec{\nabla})\vec{r}-\frac{r^2}{2}\vec{m}(\vec{\nabla}.\vec{r})$$
It is also known that $\vec{\nabla}.\vec{r}=2$.
$$\frac{r^2}{2}(\vec{m}.\vec{\nabla})\vec{r}-r^2\vec{m}$$
But I could not get desired identity.
 A: Your expressions need more parentheses than it has. Either way, let's start with the identity
$$\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B} (\vec{A} \cdot \vec{C}) - \vec{C} (\vec{A} \cdot \vec{B}).$$
Applying this to our expression gives
$$\nabla \times \left( \frac{r^2}{2} \vec{r} \times \vec{m} \right) = \vec{r} \left( \nabla \cdot \left( \frac{r^2}{2} \vec{m} \right) \right) - \vec{m} \left( \nabla \cdot \left( \frac{r^2}{2} \vec{r} \right) \right).$$
Expanding the RHS yields, for the first term,
$$\nabla \cdot \left( \frac{r^2}{2} \vec{m} \right) = \nabla \left( \frac{r^2}{2} \right) \cdot \vec{m} + \frac{r^2}{2} \nabla \cdot \vec{m} = \vec{r} \cdot \vec{m}.$$
The second term yields
$$\nabla \cdot \left( \frac{r^2}{2} \vec{r} \right) = \nabla \left( \frac{r^2}{2} \right) \cdot \vec{r} + \frac{r^2}{2} \nabla \cdot \vec{r} = \vec{r} \cdot \vec{r} + \frac{r^2}{2} \cdot \frac{2}{r} = r^2 + r.$$
This leads to
$$\nabla \times \left( \frac{r^2}{2} \vec{r} \times \vec{m} \right) = (\vec{m} \cdot \vec{r} ) \vec{r} - \vec{m}(r^2 +r).$$
For our desired expression we would need that one of the $r$'s didn't cancel, but that doesn't seem to be the case. That's because $\nabla \cdot \vec{r} = 2r^{-1}$, not just $2$.
It's not completely what we wanted, but it's pretty close. Maybe I have a calculation error somewhere.
