@cuongtd 

Not quite.

1. A single equation in 3 variables may have exactly 2 solutions. E.g.  x^2 + y^2 + z^2*(z-1)^2 = 0.
2. Inequalities do count! E.g.  x+y = 0, x >= 0, y >= 0.

If you are interested in the posted system (and not in a very general and fast method, which actually does not exist), why don't you use first f7? It simplifies the task a lot as I have mentioned.

BTW, a general method here would de to define a procedure

t1 -> fsolve(...,[p1,t2,p2,t3,p3,x,y])

The problem is that fsolve gives only one solution. But even if you can obtain all the solutions (using e.g. the avoid option, or DirectSearch) it will be very difficult to identify the curves (just think that implicitplot3d usually has difficulties, and in your problem there are 8 variables instead of 3!).

@acer 

Probably a more constructive question would be:

Given a polynomial f in one or more variables find (if possible) polynomials u,v such that
f = a*u^m + b*v^n,
a,b being constants and m,n integers.

@Thomas Richard 

Unfortunately, the .eps for 3d graphics is terrible:

plotsetup(cps,plotoutput=`d:/temp/myplot3.eps`,plotoptions=`landscape`):
plot3d(x^2-y^2,x=-1..1,y=-1..1);

Here is a printscreen (Maple 2015, 64bit, Windows 7):

That's the animation produced in Maple 2015

Edit: seems to be OK, only the viewing angle is different (and not relevant).

The animation produced after executing the worksheet is not what you claim (at least in Maple 2015.2).

@Markiyan Hirnyk 

I never do because I think that the Maplesoft members here are in a better position to decide.

in question is

C:=2*sum(1/i^4, i=1..infinity);

OP's sum:
evalf[30](sum(sum(1/(i^4+j^4),i=1..infinity),j=1..infinity) + C);
      2.92637815841768603016360338185+0.*I

@Kitonum 

OK, I just wanted to have N as parameter, without thinking at speed.

It should have been:

eval(  'add'('add'('evalf'(a), i=1..N), j=1..N),  N=1000);

I understand op, nops as reflecting the logical representation of the expression.

@John Fredsted

Your question reduces to explain:

subs(a*b=c, 7*a*b); #  7*c
subs(a*b=c, I*a*b); #  I*a*b

The answer: distinct internal representation, as shown by

dismantle(7*a*b); dismantle(I*a*b);

namely, 7*a*b has two first level operands: 7 and a*b
but I*a*b has three (I, a, b).

The unpleasant fact is that op and nops do not reflect in this case the internal representation.

My impression is that the only problem is the limited capability of assume, as the next very simple examples show.

assume(p::integer, p>1, p<4);
assume(x<3,x>2);
about(p);
   Originally p, renamed p~:
     is assumed to be: AndProp(integer,RealRange(2,3))

[floor(x)-2, floor(p/2)-1, sin(floor(p/2)*Pi), cos(floor(p/2)*Pi)+1, sin(p-2)*(p-3)];



(should be all 0).

@Preben Alsholm 

I don't think that equality is the real issue here because <= is for sure a valid condition and p<=0,p>=0 gives p=0 (and about(p) confirms this).

@Preben Alsholm 

It seems that the assume database is not always inspected (correctly?):

[p, sin(p), sin(Pi*p)] assuming p=0;
  
    [p, sin(p),0]

@Markiyan Hirnyk 

int(x^(-ln(x)), x = 0 .. b);

also works (without assumptions)!

@JacquesC 

But most of these examples can be obtained in Maple 2015 as "one-liners", e.g.

Change(int(x^(-ln(x)),x=0..b), x=ln(t)) assuming b > 0;

Edit: I wanted x=exp(t), but even this strange change of variable works!