PROBLEM: for instance to rationalize 
find which makes a




It is easily shown, that the complete problem can be solved if it can be solved for and .
0) Find ; new intermediate radicals are
1) Select the prime number p which is a factor of L, rewrite the radicals as , consider the inner roots as new radicands
2)  To get rewrite where the are replaced by in .
 To get rewrite where the are replaced by in .
Solve for the resulting expression (1 + S1a) or (1 + S1b), the new L for the remaining radicals will be L/p
Repeat for the result of (3) from (1) until no radicals remain
The 'damage' done by the 'factoring' in (2) is simple to repair.
THE I N T U I T I O N I WAS UNABLE TO PROVE UP TO NOW IS , 
For instance

I would like to know if the intuition is right ( I checked it for discrete cases as far as 1 gibabyte swap, using the computer algebra program MuPAD, allowed )
and some HINTS how to prove it.
Proving for sqareroots is easy:
If it's true for and then in the product is substituted with the x of the expanded product are replace by
and the expanded product can be written , where R,S are polynomials. Take the produkt both nonPoly values
and you again get a plolynomial, and in the product each factor f is substituted by the two factors .