<s>[INST] prove: $\nabla \cdot \frac{\vec{r}}{r^3} = 0$ prove:
$\nabla \cdot \frac{\vec{r}}{r^3} = 0$

a few useful identities for proof:
$\vec{r} \cdot \vec{r} = 1$
$\nabla r^n = n r^{n-2} \vec{r}$
$\nabla \cdot (\vec{v} + \vec{w}) = \nabla \cdot \vec{v} + \nabla \cdot \vec{w}$
$\nabla \cdot (A \vec{v}) = (\nabla A) \cdot \vec{v} + A (\nabla \cdot \vec{v})$
$r = (x^2 + y^2 + z^2)^{-\frac{1}{2}}$
$\vec{r} = x\hat{\text{i}} + y\hat{\text{j}} + z\hat{\text{k}}$

ok...so i attempt this proof as follows:
$\nabla \cdot \frac{\vec{r}}{r^3} = \nabla ( r^{-3}~ \vec{r})$
by product rule:
$\nabla \cdot \frac{\vec{r}}{r^3} = (\nabla r^{-3}) \cdot \vec{r} + r^{-3}(\nabla \cdot \vec{r})$
by one of the identities that kind of looks like a derivative rule:
$\nabla \cdot \frac{\vec{r}}{r^3} = -3r^{-5}\vec{r} \cdot \vec{r} + r^{-3}(\nabla \cdot \vec{r})$
then because $\vec{r} \cdot \vec{r} = 1$:
$\nabla \cdot \frac{\vec{r}}{r^3} = -3r^{-5} + r^{-3}(\nabla \cdot \vec{r})$
At this point i get stuck trying to get this expression to equal zero...
 [/INST] Solution : $$\nabla \cdot\frac{\vec r}{r^3}$$
$$=(\nabla \cdot \vec r)\frac{1}{r^3} + \vec r \cdot \left(  \nabla \frac{1}{r^3} \right)$$
where, $\nabla \cdot(\phi \vec A)=(\nabla \cdot \vec A) \phi + \vec A \cdot (\nabla \phi)$
$$=\frac{3}{r^3}+\vec r \cdot [ (-3)r^{-3-2}\vec r ] $$
where , $ \nabla r^n=nr^{n-2} \vec r$
$$=\frac{3}{r^3}+\frac{-3}{r^3}=0$$
</s>