I propose a different proof of this remarkable identity (see  http://www.mapleprimes.com/posts/144499-Stunningly-Beautiful-Identity-Proved ) in which  directly constructed a polynomial, whose root is the value of LHS, and this is expressed in radicals.

For the proof, we need three simple identities with cubic roots (a, b, c -any real numbers):

restart;

is((a^(1/3)+b^(1/3)+c^(1/3))^3=3*(a^(1/3)+b^(1/3)+c^(1/3))*(a^(2/3)+b^(2/3)+c^(2/3))-2*(a+b+c)+6*a^(1/3)*b^(1/3)*c^(1/3));

is(a^(2/3)+b^(2/3)+c^(2/3)=(a^(1/3)+b^(1/3)+c^(1/3))^2-2*(a^(1/3)*b^(1/3)+a^(1/3)*c^(1/3)+b^(1/3)*c^(1/3)));

is((a^(1/3)*b^(1/3)+a^(1/3)*c^(1/3)+b^(1/3)*c^(1/3))^3=(a*b+a*c+b*c)+6*a^(2/3)*b^(2/3)*c^(2/3)+3*a^(1/3)*b^(1/3)*c^(1/3)*(a^(1/3)+b^(1/3)+c^(1/3))*(a^(2/3)+b^(2/3)+c^(2/3))-3*(a+b+c)*a^(1/3)*b^(1/3)*c^(1/3));

                                                        true

                                                        true

                                                        true

Define constants:

a:=cos(3*Pi/19)+cos(5*Pi/19)+cos(17*Pi/19):

b:=cos(Pi/19)+cos(7*Pi/19)+cos(11*Pi/19):

c:=cos(9*Pi/19)+cos(13*Pi/19)+cos(15*Pi/19):

Checking  of another 3  identities:

simplify([a+b+c, a*b*c, a*b+a*c+b*c]);

                    [1/2, -7/8, -3/2]

Denote by  x = a^(1/3)+b^(1/3)+c^(1/3),  y = a^(2/3)+b^(2/3)+c^(2/3),  z = a^(1/3)*b^(1/3)+a^(1/3)*c^(1/3)+b^(1/3)*c^(1/3)  (here a, b, c are symbols)  . 

Using the above identities make up the system and solve it:

solve({x^3=3*x*y-1-3*7^(1/3), y=x^2-2*z, z^3=-3/2+6*7^(2/3)/4-3*7^(1/3)/2*x*y+3*7^(1/3)/4});

We are only interested in the value of  x :

[allvalues(rhs(%[1]))]:

Of 9 roots of this polynomial, only the first root is a real number. This will be the value in the radicals of a^(1/3)+b^(1/3)+c^(1/3):

Check that this  number equals the number RHS:

is(%[1]=(1/2-3*7^(1/3)+(3/2)*(-25+3*7^(2/3)+18*7^(1/3))^(1/3)+(3/2)*(-44+18*7^(1/3)+3*7^(2/3))^(1/3))^(1/3));

                                                             true

Trigonometric_identi.mws