Prove that $$\nabla.\left(\vec{r}\,\nabla\left(\frac{1}{r^3}\right)\right)= \frac3{r^4}.$$
My doubt is that here should we consider the term $\vec{r}$ and $\nabla\left(\frac{1}{r^3}\right)$ as the ones to be multiplied or as dot product? Also please give the solution. Thanks in advance!
$\nabla\big(\frac{1}{r^3}\big) = \frac{-3}{r^4}\nabla(r) = \frac{-3}{r^5}\vec{r} $
$\bigg[r^2 = x^2 +y^2 + z^2$
$\frac{\partial r}{\partial x} = \frac{x}{r}$, similarly for y and z.
$\nabla r = \Sigma(\hat i \frac{\partial r}{\partial x}) = \frac{1}{r}\Sigma(x \hat i) = \frac{\vec{r}}{r}\bigg] $
$y = \nabla.\bigg( r.\nabla\big(\frac{1}{r^3}\big)\bigg) = \nabla.\bigg( r.\frac{-3}{r^5}\vec{r}\bigg) = -3\nabla.\bigg( \frac{\vec{r}}{r^4}\bigg) = -3\bigg[\frac{1}{r^4}\nabla.\bigg( \vec{r}\bigg) + \vec{r}.\nabla(\frac{1}{r^4})\bigg]$
$y =-3\bigg[\frac{1}{r^4}3 + \vec{r}.\frac{-4}{r^6}\vec{r}\bigg] = -3\bigg[\frac{1}{r^4}3 + {r^2}\frac{-4}{r^6}\bigg] = \frac{3}{r^4} $
[Formulae used: $grad(r) = \frac{\vec{r}}{r} \ , \ div(\vec{r}) = 3 \ , \ \vec{r}.\vec{r} = r^2 $]