Laplacian(F) = (n-1/r)g'(r) + g''(r) I got one more problem from my self reading of Methods of Advanced Calculus by Edwards, hints and solutions are equally appreciated:
If f(x) = g(r), r= |x|, and n>=3, show that 
Laplace(f) = d^2f/dx1^2 +.....+d^2f/dx2^2 = (n-1/r)g'(r) + g''(r)
and using this result show that if Lapace(f) =0 then f(x) = a/|x|^n-2 + b when x does not = 0
Does the n-1 come from the definition of the derivative ?

$r^2 = ||<x_1,x_2, \dots x_n> ||^2 = x_1^2+x_2^2+ \cdots x_n^2$. Consequently,
$ r \frac{\partial r}{\partial x_j} = x_j $. Consider then, by the chain-rule and the result above:
$$ (\nabla g)_j  = \frac{\partial }{\partial x_j} g(r) = g'(r)\frac{\partial r }{\partial x_j} = g'(r) \frac{x_j}{r}$$
Now, differentiate once more:
$$ \frac{\partial}{\partial x_j}(\nabla g)_j  = \frac{\partial }{\partial x_j} \left( g'(r) \frac{x_j}{r} \right) = g''(r)\frac{x_j^2}{r^2}+\frac{\partial }{\partial x_j}\frac{x_j}{r} = g''(r)\frac{x_j^2}{r^2}+ g'(r)\frac{r-x_j\frac{x_j}{r}}{r^2}$$
cleaning it up a bit:
$$ \frac{\partial}{\partial x_j}(\nabla g)_j= g''(r)\frac{x_j^2}{r^2}+ g'(r)\frac{r^2-x_j^2}{r^3}$$
Finally, to find the Laplacian sum the formula above from $j=1,2, \dots n$. It's obvious where the $g''(r)$ term arises. On the other hand, you get $n$ copies of $r^2$ less one copy of $r^2$ from the $g'(r)$ term. Think on this for a bit, you'll see it. Welcome to the MSE.
