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Question: how to obtain this simplification using Maple commands?

Question:how to obtain this simplification using Maple commands?

nm 9934 Maple
solve eliminate
August 19 2019
1

From a book, it shows the following

Verified by hand the last result  above x^(2/3)+y^(2/3)-a^(2/3)=0 is correct. The input is always 2 equations in x and y as shown above, and there is always one constant C in both that needs to be eliminated to obtain a solution (one equation) that contains y,x and any other parameters, but without c.

have been trying to use eliminate command to do the same as above. I assume eliminate is the right command for this. But not able to get close to what the book shows above for final result. 

Does any one knows how obtain same result as above using Maple's eliminate?  (I can't follow the same steps as hand solution, since that would apply only to the above example. I need to use a generic approach). 

Sometimes it is hard to obtain same result using computer as one can do by "hand".

Here is some of my attempts

> 

restart;
> 

assume(x::real,y::real,a::real);
eq1:=x=-a/(1+c^2)^(3/2);
eq2:=y=a*c^3/(1+c^2)^(3/2);

x = -a/(c^2+1)^(3/2)

y = a*c^3/(c^2+1)^(3/2)

> 

result:=eliminate([eq1,eq2],c);
result:=DEtools:-remove_RootOf(result[2,1]):
result:=DEtools:-remove_RootOf(result);
result:=simplify(result,power,symbolic);
result:=expand(result);

[{c = RootOf(_Z^2-RootOf(_Z^3*x+a)^2+1)}, {y*(RootOf(_Z^3*x+a)^2)^(3/2)-RootOf(_Z^2-RootOf(_Z^3*x+a)^2+1)*RootOf(_Z^3*x+a)^2*a+a*RootOf(_Z^2-RootOf(_Z^3*x+a)^2+1)}]

x*((a^2-y^2)*(a*y^2)^(1/3)*a*((a*y^2)^(2/3)+a*(a*y^2)^(1/3)+y^2))^(3/2)/(a*y^2*(a^2-y^2)^3)+a = 0

a*((y^(2/3)*a^(2/3)+a^(4/3)+y^(4/3))^(3/2)*x+(a^2-y^2)^(3/2))/(a^2-y^2)^(3/2) = 0

x*a*(y^(2/3)*a^(2/3)+a^(4/3)+y^(4/3))^(3/2)/(a^2-y^2)^(3/2)+a = 0

 

Download how_to_eliminate.mw

Notice: The reason I am asking the above, is becuase I was doing it this way: I first solve for c from one equation, then use this result in the second equation (this is what one would do normally by hand). but this could result in many solutions and hard to know which to pick to match the book result. That is why I am thinking of using Elminate instead:

> 

restart;
> 

assume(x::real, y::real,a::real);
eq1:=x=-a/(1+c^2)^(3/2);
eq2:=y=a*c^3/(1+c^2)^(3/2);

x = -a/(c^2+1)^(3/2)

y = a*c^3/(c^2+1)^(3/2)

> 

#brute force method
c_found:=Vector([solve(eq1,c)])

Vector(6, {(1) = sqrt((-a*x^2)^(2/3)-x^2)/x, (2) = -sqrt((-a*x^2)^(2/3)-x^2)/x, (3) = (1/2)*sqrt(2)*sqrt(I*sqrt(3)*(-a*x^2)^(2/3)-(-a*x^2)^(2/3)-2*x^2)/x, (4) = -(1/2)*sqrt(2)*sqrt(I*sqrt(3)*(-a*x^2)^(2/3)-(-a*x^2)^(2/3)-2*x^2)/x, (5) = (1/2)*sqrt(-(2*I)*sqrt(3)*(-a*x^2)^(2/3)-2*(-a*x^2)^(2/3)-4*x^2)/x, (6) = -(1/2)*sqrt(-(2*I)*sqrt(3)*(-a*x^2)^(2/3)-2*(-a*x^2)^(2/3)-4*x^2)/x})

> 

map(x->simplify(subs(c=x,eq2),symbolic),c_found)

Vector(6, {(1) = y = -(1/4)*sqrt(2)*(I*a^(2/3)*sqrt(3)-a^(2/3)-2*x^(2/3))^(3/2), (2) = y = (1/4)*sqrt(2)*(I*a^(2/3)*sqrt(3)-a^(2/3)-2*x^(2/3))^(3/2), (3) = y = -((1/4)*I)*sqrt(2)*(I*a^(2/3)*sqrt(3)+a^(2/3)+2*x^(2/3))^(3/2), (4) = y = ((1/4)*I)*sqrt(2)*(I*a^(2/3)*sqrt(3)+a^(2/3)+2*x^(2/3))^(3/2), (5) = y = -(a^(2/3)-x^(2/3))^(3/2), (6) = y = (a^(2/3)-x^(2/3))^(3/2)})

> 

 

 

Looking at result above, I think I can safely eliminate all y solutions with complex number I in them. This leaves the last two listed above (real y). Which is a little better than before.

Download how_to_eliminate_brute_force.mw

 

 
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