@Markiyan Hirnyk 
The questioner wrote "two positive solutions and two negative solutions". If  m=5/2  then we have one negative and two positive roots.

@Markiyan Hirnyk 
The questioner wrote "two positive solutions and two negative solutions". If  m=5/2  then we have one negative and two positive roots.

@Markiyan Hirnyk 

Of course, I understand that the field is not potential, so the path of integration should be specified.

@Markiyan Hirnyk 

Of course, I understand that the field is not potential, so the path of integration should be specified.

This is a simple task, it is easy to solve by a procedure. But I did not understand the law of getting your matrix for any m.  What is  i ? Can you give an example of such a matrix for a given m, for example, for m = 8 ?

I noticed an error in the text of the procedure. Instead of  if alpha = 0 then {k> 0}  should be  if alpha = 0 then print(`No solutions`) .

@Markiyan Hirnyk

Of course, the advantage of substantially purely aesthetic.
Also, I think this form of writing can be interesting for educational purposes in the study of the concept of absolute value.

PS. The advantage is the same as   |x|  over 

I did not notice the condition that  x>=0, y>=0, z>=0. Therefore, the solution exists.

I did not notice the condition that  x>=0, y>=0, z>=0. Therefore, the solution exists.

@one_man

Thank you for your interest! I was waiting for your question, because I know about your enthusiasm implicitly given curves. Setting the boundary of the figure is not provided in this way in my procedures, but still there is a simple solution to the problem. Here it is: 

P1 := plots[implicitplot]((x1+0.7e-1*sin(30*x2)^2)^2+(x2+0.7e-1*sin(30*x1)^2)^2-1 = 0, x1 = -2 .. 2, x2 = -2 .. 2, numpoints = 100000):

P2 := plottools[getdata](P1):

L := [convert(P2[3], listlist)]:

Perimeter(L);

Area(L);

plots[display](Picture(L, color = green, [color = black, thickness = 2]));

@Alejandro Jakubi 

Of course, my statement is not accurate from a mathematical point of view. Would be more accurate to write that there are examples which algsubs command decides better than applyrule command. Maybe you know the opposite examples?

Thanks for your comment!

@Alejandro Jakubi 

Of course, my statement is not accurate from a mathematical point of view. Would be more accurate to write that there are examples which algsubs command decides better than applyrule command. Maybe you know the opposite examples?

Thanks for your comment!

I didn't realize what  P[0] and  plotofmaxmin  stand for?

@Axel Vogt 

If we set

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):

then with Maple easily find

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

                    [1/2, -7/8, -3/2]

This result gave me the idea to associate in a single system known symmetric expressions  a+b+c, a*b*c, a*b+a*c+b*c  and unknown  expressions  a^(1/3)+b^(1/3)+c^(1/3),   a^(2/3)+b^(2/3)+c^(2/3)   and   a^(1/3)*b^(1/3)+a^(1/3)*c^(1/3)+b^(1/3)*c^(1/3) .  

To make up such a system, I used three identities, which are valid for any reals

(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),

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)),

(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)

Firstly I got these identities by hand and then checked in Maple. They based on the theory of symmetric polynomials. It is known that any symmetric polynomial can be expressed as a polynomial in the elementary symmetric polynomials. My first two identities follow from the expressions

x^3 + y^3 + z^3 = 3*x*y*z + (x + y + z)^3 - 3 (x + y + z)*(x*y + x*z + y*z)  and

x^2 + y^2 + z^2 = (x + y + z)^2 - 2*(x*y + x*z + y*z) 

in which it was substituted

x = a^(1/3) ,  y = b^(1/3) ,  z = c^(1/3)   and

the third identity is obtained by hand through expansion of

 (a^(1/3)*b^(1/3)+a^(1/3)*c^(1/3)+b^(1/3)*c^(1/3))^3

Carl Love! 

Thanks for the elegant proof. I wonder whether can your approach  find the value of LHS in the real radicals if RHS is unknown?

I propose a different proof of this remarkable identity in which  directly constructed a polynomial, whose root is the value of LHS, and this is expressed in radicals. As this approach is completely unrelated to your method, I am creating a separate post.