@acer 

 Excellent, these clever & helpful suggestions work to solve my problem.  Thank you very much !

@acer 

Yes, this is going in the direction that I need: removing all non-math garble.  Thank you very much !!

@acer 

Yes,  for the reasons commented above, the output of ifactor is hard to use in further work. This is why I think that a proper implementation of routines for dealing with decomposition of numbers would be a nice addition to Maple. Algebra is good, arithmetics is missing.

@Carl Love 

Hi, I need to extract all bases because to the integer of departure  there is an associated polynomial that needs to be decomposed.

Such decompositions involve coefficients which are non-quadratic algebraic numbers. So, I need all bases to be tested

as Ansatz for decompositions.  While Maple has friendly routines to deal with decomposing polynomials, the analogous

handling of integer numbers does not seem to have been implemented yet. This would be a handy thing to add. Note that

even the list of numbers produced by ifactors looks odd, withsign of the integer being decomposed is written "1", instead of  [-1,0], for example. as all the other factors are written.

For 1125 = 3^2*5^3, a simple polynomial is a x^4-x^3-4*x^2+4*x+1  (Galois  4T1) which Maple breaks using the integer with odd exponent:    factor(x^4-x^3-4*x^2+4*x+1, 5^(1/2) );   [5 has the odd exponent]

However, for 725 = 5^2 * 29, the associated 4T3 polynomial is x^4-x^3-3*x^2+x+1, not decomposable using

factor(x^4-x^3-3*x^2+x+1,  29^(1/2) ), but with  factor(x^4-x^3-3*x^2+x+1, 3^(1/2) ),  [3 has the even exponent].

Only cyclic  4T1 polynomials will decompose "easily", using bases with odd exponents. For quartics one may of course use solve to get all 4 roots. But simple quartics were used here just to explicitly show the origin of complications that Maple could be more helpful to deal with. I hope this all to be clear.  Thank you.

@acer 

Thank you !!

 Is there a simple way to get rid of the initial  "1"  and  all subsequent nested parenthesis?

  Thank you very much for your great help!

   # monogenic sextics, with (minimal) discriminant, 300125:
   mono:=Matrix(3,2):  
   mono[1,1] := x^6+x^5-7*x^4-2*x^3+7*x^2+2*x-1:
   mono[1,2] := x^6+3*x^5-6*x^4-10*x^2+18*x+1:
   mono[2,1] := x^6+3*x^5-6*x^4-10*x^3+18*x^2-8*x+1:
   mono[2,2] := x^6+5*x^5+3*x^4-16*x^3-24*x^2-9*x-1:
   mono[3,1] := x^6+4*x^5-2*x^4-26*x^3-34*x^2-13*x-1:
   mono[3,2] :=x^6+9*x^5+24*x^4+2*x^3-66*x^2-40*x+41:
   Lat(mono);

I correctly get:

 \begin{tabular}{|c|c|} \hline ${x}^{6}+{x}^{5}-7\,{x}^{4}-2\,{x}^{3}+7\,{x}^{2}+2\,x-1$ & ${x}^{6}+3\,{x}^{5}-6\,{x}^{4}-10\,{x}^{2}+18\,x+1$ \\ \hline ${x}^{6}+3\,{x}^{5}-6\,{x}^{4}-10\,{x}^{3}+18\,{x}^{2}-8\,x+1$ & ${x}^{6}+5\,{x}^{5}+3\,{x}^{4}-16\,{x}^{3}-24\,{x}^{2}-9\,x-1$ \\ \hline ${x}^{6}+4\,{x}^{5}-2\,{x}^{4}-26\,{x}^{3}-34\,{x}^{2}-13\,x-1$ & ${x}^{6}+9\,{x}^{5}+24\,{x}^{4}+2\,{x}^{3}-66\,{x}^{2}-40\,x+41$ \\ \hline \end{tabular}