<s>[INST] 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.
 [/INST] $$
\nabla^2r^3=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2})(x^2+y^2+z^2)^{3/2}\\
= \frac{\partial }{\partial x }(3x(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial y }(3y(x^2+y^2+z^2)^{1/2})+\frac{\partial }{\partial z }(3z(x^2+y^2+z^2)^{1/2}) \\
=\frac{3(2x^2+y^2+z^2)}{(x^2+y^2+z^2)^{1/2}}+\frac{3(x^2+2y^2+z^2)}{(x^2+y^2+z^2)^{1/2}} +\frac{3( x^2+y^2+2z^2)}{(x^2+y^2+z^2)^{1/2}}\\=3\frac{ (4 x^2+4y^2+4z^2)} {(x^2+y^2+z^2)^{1/2}} =12r
$$
$$
\nabla\cdot\big[\frac{(x, y, z)}{(x^2 + y^2 + z^2)^{p/2}}\big] =\frac{3-p}{(x^2+y^2+z^2)^{p/2}}=\frac{3-p}{r^{p }}
$$
So $\nabla\cdot\frac{{\bf r}}{r^p}=0$ when $p=3$.
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