Proving Vector Identities Let r=$(x,y,z)$ and $r=$||r||.
(A) Prove that $\nabla^2r^3=12r$.
(B) Is there a value of $p$ for which the vector field f(r) = r/$r^p$ is solenoidal?
What I have tried:
For part (a) I think that $r^3=\sqrt{27} $ but I am unsure of what to do next in terms of the del operator.
 A: $$
\nabla^2r^3=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})(x^2+y^2+z^2)^{3/2}\\
= \frac{\partial }{\partial x }(3x(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial y }(3y(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial z }(3z(x^2+y^2+z^2)^{1/2}) \\
=\frac{3(2x^2+y^2+z^2)}{(x^2+y^2+z^2)^{1/2}}+\frac{3(x^2+2y^2+z^2)}{(x^2+y^2+z^2)^{1/2}} +\frac{3( x^2+y^2+2z^2)}{(x^2+y^2+z^2)^{1/2}}\\=3\frac{ (4 x^2+4y^2+4z^2)} {(x^2+y^2+z^2)^{1/2}} =12r
$$
$$
\nabla\cdot\big[\frac{(x, y, z)}{(x^2 + y^2 + z^2)^{p/2}}\big] =\frac{3-p}{(x^2+y^2+z^2)^{p/2}}=\frac{3-p}{r^{p }}
$$
So $\nabla\cdot\frac{{\bf r}}{r^p}=0$ when $p=3$.
A: A more concise approach to (A) uses$$\nabla^2r^3=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}r^3\right)=\frac{1}{r^2}\frac{d}{dr}\left(3r^4\right)=12r.$$Similarly, for (B) use$$\nabla(r^{-p})=\frac{d}{dr}(r^{-p})\hat{\mathrm{r}}=-pr^{-p-2}\mathrm{r}$$to deduce$$\nabla\cdot(r^{-p}\mathrm{r})=r^{-p}\nabla\cdot\mathrm{r}+\nabla(r^{-p})\cdot\mathrm{r}=(3-p)r^{-p},$$i.e. the field is solenoidal iff $p=3$ (which would surprise no physicists familiar with the differential form of Gauss's law).
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\large\pars{A}:}$
\begin{align}
\nabla^{2}r^{3} & = \nabla\cdot\pars{\nabla r^{3}} =
\nabla\cdot\pars{\totald{r^{3}}{r}\,{\vec{r} \over r}} =
3\nabla\cdot\pars{r\,\vec{r}}
\\[5mm] & =
3\bracks{\pars{\nabla r}\cdot\vec{r} + r\nabla\cdot\vec{r}}
\\[5mm] & =
3\bracks{\pars{\totald{r}{r}\,{\vec{r} \over r}}
\cdot\vec{r} + r\pars{3}}
\\[5mm] & = 3r + 9r = \bbx{12r} \\ &
\end{align}

$\ds{\large\pars{B}:}$
\begin{align}
\nabla\cdot\pars{\vec{r} \over r^{p}} & =
\pars{\nabla r^{-p}}\cdot\vec{r} +
r^{-p}\,\,\nabla \cdot\vec{r}
\\[5mm] & =
\pars{\totald{r^{-p}}{r}\,\,{\vec{r} \over r}}\cdot\vec{r} + r^{-p}\pars{3}
\\[5mm] & =
\underbrace{\pars{-pr^{-p - 1}\,\,\,{\vec{r} \over r}}\cdot\vec{r}}_{\ds{-pr^{-p}}}\
+\ 3r^{-p}
\\[5mm] & =
\pars{-p + 3}r^{-p}
\end{align}
Then,
$$
\nabla\cdot\pars{\vec{r} \over r^{p}} = 0 \implies
\bbx{p = 3} \\
$$
