@Markiyan Hirnyk 

Just to prevent nonsense inputs:

rad := proc(u::polynom) `*`(op(map( z->op(1,z),factors(u)[2]))) end:

@Markiyan Hirnyk 

The ring of the polynomials, of course, K[x,y,...], K an extension of Q.

@Markiyan Hirnyk 

1. The results are the same, modulo the units of the ring.
One may apply primpart to all the terms, but it is not necessary.

2. You are right. But over the reals, factors works better.
And if we consider e.g. polynomials over a noncommutative field, then both methods fail. :-)

@Markiyan Hirnyk 

I lnow that it may fail; actually it could be fixed.

The real point was that factor could work where Radical fails.

@Markiyan Hirnyk 

Radical also fails for such polynomials (with irrational coeffs).

factor may work. E.g.

g:=expand((x*sqrt(2)+y*sqrt(3)+z*sqrt(6)+x*y*sqrt(6)+x)^6):
r:=factor(g):
if op(0,r)=`*` and type(op(1,r),realcons) then r:=op(2,r) fi:
if op(0,r)=`^` then op(1,r) else FAIL fi;

@Carl Love 

As mentioned It works in the more general case when {op(L2)} is a subset of {op(L1)},
so o permutation may not exist, e.g. [1,2,3] --> [1,2,1].

@I_Mariusz 

I'm also waiting for a version where

is(Re(RootOf(Zeta(z),z))=1/2) assuming Re(z)>0, Re(z)<1;

will work. Or, even better:

is(Re(RootOf(Zeta(z),z))=1/2, option=proof) assuming Re(z)>0, Re(z)<1;

@Kitonum 

I think that a simple simplify should produce

k1*(.1124471606*n^2*k1^2-.2042283048*k2*n^2*k1-0.1874763818e-1*k2^2*n^2+.2042283048*k2*n*k1
+.1025985806*k1*n^2+0.3749527637e-1*k2^2*n+.1349080187*k2*n^2-0.4685942835e
-1*k1*n-0.1874763818e-1*k2^2-.2698160373*k2*n-.1823825149*n^2-0.5573915226e-1*k1
+.1349080187*k2+.3647650297*n-.1823825149)/(n-1.)

To obtain this it seems that we must enter five commands:

normal(evalf(expand(normal(convert(A,rational)))));

@Markiyan Hirnyk 

I also suspect that FindFormula will try to be for functions what Maple's identify is for numbers.
I.e. identify will act as FindFormula applied to a constant input.
I wonder how accurate will/could it be!

@Preben Alsholm 

A related question about searching the help system

Say we are searching for inert function.
The most relevant entry would be ?value
But in the help system, 'value' appears as the 20th entry.
I also find that the (rather new) organisation of the Table of Contents
in help system is less intuitive.

What keywors should we use instead of inert function
such that "value" appears in the first place?
(except for value, of course).

For now, it seems that a google search is more effective!
For example,

maple inert function

sends us directly to
http://www.maplesoft.com/support/help/Maple/view.aspx?path=value

i.e. the best place!

@EugeneKalentev 

A box is determined by xmin,ymin,zmin,xmax,ymax,zmax.  E.g.

B:=Box(-1..1,-1..1,-1..1, outward);

is the box having the faces tangent to your sphere.

@Carl Love 

The simplest is

int(1/sqrt(x),x=0..1);

After x=t^2 ==> int(2,t=0..1)

@Carl Love 

Mathematically, the flux is a surface integral; it reduces to a double integral. If the integral exists (i.e. the integrand is an integrable function in any sense - Riemann/improper/Lebesgue etc) then there is no math problem. 

In this case it exists. Note that many improper integrals become proper after a change of variables.

The divergence theorem was not used here; and it cannot be used. Actually Div(vv) is not integrable in any ball centered at <0,0,0>.

@acer 

The  "By hand" method is the easiest for a beginner: simply copy the output of Flux(...,inert) and paste it in a new execution group. Then swap  phi = 0 .. Pi    and  theta = 0 .. 2*Pi.
(with no escaped variables problem).

@EugeneKalentev 

That is exactly what evalf does if applied to an inert integral (see ?int/details): the integral is computed numerically even if a closed form exists.