<s>[INST] Let $\mathbf{r}=(x,y,z)$,$r=||\mathbf{r}||$. Show the following equation on $B\cdot \nabla (A\cdot \nabla (\frac{1}{r}))$ Let $\mathbf{r}=(x,y,z)$ and let $r=||\mathbf{r}||$. If $A$ and $B$ are constant vectors show that:
$$B\cdot \left(\nabla \left (A\cdot \nabla \left(\frac{1}{r}\right)\right)\right)=\frac{3A\cdot \mathbf{r}B\cdot \mathbf{r}}{r^5}-\frac{A\cdot B}{r^3}$$
I've found so far that $A\cdot \nabla\left(\frac{1}{r}\right)=-\frac{A\cdot \mathbf{r}}{r^3}$.
However, I have not been able to show the equation above. 
I would greatly appreciate any solutions, suggestions, or hints.
 [/INST] Let's start from what you have found so far, $A\cdot \nabla(\frac{1}{r})=-\frac{A\cdot \mathbf{r}}{r^3}$, and denote vector constant vector $A$ as $(a_x,a_y,a_z)$. The expression can be re-written as
$$A\cdot \nabla\left(\frac{1}{r}\right)=-\frac{A\cdot \mathbf{r}}{r^3}=-\frac{a_xx+a_yy+a_zz}{(x^2+y^2+z^2)^{3/2}}$$ 
Evaluating the partial derivative with respect to $x$, $y$ and $z$ (using the product rule of differentiation for each partial derivative) results in
$$\begin{align}\frac{\partial}{\partial x}A\cdot \nabla\left(\frac{1}{r}\right)&=\frac{3x(a_xx+a_yy+a_zz)}{(x^2+y^2+z^2)^{5/2}}-\frac{a_x}{(x^2+y^2+z^2)^{3/2}}\\
\frac{\partial}{\partial y}A\cdot \nabla\left(\frac{1}{r}\right)&=\frac{3y(a_xx+a_yy+a_zz)}{(x^2+y^2+z^2)^{5/2}}-\frac{a_y}{(x^2+y^2+z^2)^{3/2}}\\
\frac{\partial}{\partial z}A\cdot \nabla\left(\frac{1}{r}\right)&=\frac{3z(a_xx+a_yy+a_zz)}{(x^2+y^2+z^2)^{5/2}}-\frac{a_z}{(x^2+y^2+z^2)^{3/2}}\end{align}$$
Combining all three partial derivatives leads to
$$\nabla (A\cdot \nabla\left(\frac{1}{r}\right))=\frac{3(A\cdot\mathbf{r})\mathbf{r}}{r^5}-\frac{A}{r^3}$$
The final stage is more straightforward, leading to the final result:-
$$B\cdot \nabla (A\cdot \nabla (\frac{1}{r}))=B\cdot \left(\frac{3(A\cdot\mathbf{r})\mathbf{r}}{r^5}-\frac{A}{r^3}\right)=\frac{3(A\cdot\mathbf{r})(B\cdot\mathbf{r})}{r^5}-\frac{A\cdot B}{r^3}$$
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