@Carl Love I was working with Groebner Basis and the result suggested symmetric polynomials as basis. To my surprise I found cyclic polynomials among them and also the fact that Maple has no refrences for these two classes of polynomials (is this correct?)

Anyhow. The case is 3 variable case and so I find cyclic polynomials can be reduced to a linear combination of cyclic polynomials of the form x^n*y+y^n*z+z^n*x, coefficients in symmetric polynomials.

@vv I meant algebraic relationships akin to those of symmetric polynomials. I take it matrix operations are unavoidable.

@vv Thank You. Are you saying this function relates that webpage?

But what are you simplifying?

You're meant to present the file on the forum. Use the green arrow to upload the worksheet.

@mmcdara I apologize. In my haste it seems I have not included the requirement on the generators to be mutually disjoint. Minimal base would have least count. You can always subdivide non trivial basis elements, but that would increase the count.

Also I should point out that intersection and union are both n-ary operators, rather than simply binary. Every combination of sets is considered for intersection.

@Carl Love I believe so. It would be a minimal basis of disjoint elements, such that every intersection of elements of S is presentable as a union of elements of said basis.

An example:   F({{1, 2, 3}, {1, 4, 5}, {2, 3, 4}}) = {{}, {1}, {4}, {5}, {2, 3}}; and F({{1, 2, 3, 6}, {1, 4, 5, 6}, {2, 3, 4, 6}}) = {{1}, {4}, {5}, {6}, {2, 3}};

So it seems I forgot to include these criterions for T:
3. every intersection of elements of S is presentable as a union of elements of T uniquely
4. T is minimal

(I think the above means we can skip 1.)

Instead of removing my own answer, I accidentally removed someone else's.

You have already spotted the essential difference. The unevaluate is needed when using the $.

@goli Apply this code to your expression:

A := RootOf(6*_Z^3+(27+3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z^2+(3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2-9*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+90*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-18*l^4+6*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2-81+45*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2))*_Z-324-3*l^8+l^10*RootOf(_Z^2*l^2+3*_Z^4-3)^2+108*RootOf(_Z^2*l^2+3*_Z^4-3)^2*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)-3*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^6+sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^8*RootOf(_Z^2*l^2+3*_Z^4-3)^2-63*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^2+30*sqrt(9-3*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2)*l^4*RootOf(_Z^2*l^2+3*_Z^4-3)^2+45*l^6*RootOf(_Z^2*l^2+3*_Z^4-3)^2+351*RootOf(_Z^2*l^2+3*_Z^4-3)^2*l^2-108*l^4, index = 1):
L := proc(A, lambda)
  collect(
  (`@`(numer, ((L) -> map2(op, 1, op(2, L))), evala,  Factors, evala, Norm))(lambda - convert(A, RootOf)), lambda)
end proc(A, lambda);
F := map(unapply, L, lambda);
factor(evala(F(A))):
subs(l = 1/2, %):
map(Testzero @ proc(x) evalf(x, 60); fnormal(%); end, %);
if member(true, %,'i') then L[i] else FAIL fi;

How effective is series approximation?

@Carl Love Visibly, one more definition is required to solve numerically...

@Carl Love Unexpected. I get unevaluated with maple 2017.

@Glowing Obviously, integers have infinite precision. Your "error" in the code is to use "1014" instead of "1014.".

@Axel Vogt 

restart;
 P1 := 1007;
 P2 := 1014;
 P3 := 1014.1;
 P2 - P1;
 P3 - P1;
 evalf(P2 - P1, 2);
 evalf(P3 - P1, 2);
 
 restart;
 P1 := 1007.;
 P2 := 1014.;
 P3 := 1014.1;
 P2 - P1;
 P3 - P1;
 evalf(P2 - P1, 2);
 evalf(P3 - P1, 2);

@taro You can replace identical(w^sigma) with identical(w)^Non(integer) as convert/parfrac will not accept non integer powers of the variable. Also, You are correct. If You negate the first element in the list then you will see the type of which  occurences will be substituted. In this case it is Non(Non(identical(w^sigma))) = identical(w^sigma). The items in the second element in the list will present without substitution, preserved. Note, the substitutions are applied only to the face of the expression passed and the result returned. If you want to see how it works out You may want to use print:

frontend(proc(E,a,v) print('args'); convert('args'); end ,[e_n_1b2,parfrac, w^sigma],[{Non}(identical(w^sigma)), {}]);