@ahmeng  Above, I showed only one solution (in my opinion, the most direct and reliable). Here are 2 more ways:

Download more_ways.mw

@ibrahimcbx 

1. You can do this with either separate commands or with a single command.

2. I didn't understand the meaning of your second question.

@Carl Love  It seems that your solution will be correct only in the segment  p in [0, 3-2*sqrt(2)]

@salma951  I didn't notice that 3 appears twice and 6 is missing. So write  randperm([1,2,3,3,4,5]) 

@jalal  Replace  range = -3 .. 10  by  (for example)  range = 1 .. 10  .

@acer  I think in all intermediate calculations it is better to use exact mathematics. Of course, we can present the final results in any desired form.

For example:

[solve({eqy3, x>=-5, x<=5})];

nm 4588  We can easily generalize your method to other exponential functions (in addition to exp). We can also use the  applyop  command (many shorter if the position of the corresponding terms is known):

restart;
expr := -2*exp(x)*x - exp(x) + 2*exp(x)^2*x + exp(x)^2 +(3^x)^2+ sin(x)^3 + cos(x)^2;
subsindets(expr,{exp(anything)^anything,(anything^anything)^anything},combine); # The first way
applyop(combine,{$1..5},expr); # The second way

@jalal  It looks even better if the multiplication sign is implied, but not displayed:

4%/5*``(2*x-5);

@nm  assuming positive greatly limits the scope of this identity, and if this condition is violated, the identity may be incorrect. For example,  -cos(A) - sin(A) <> sqrt(2)*sin(A+Pi/4)

@jalal 

restart;
randomize():
do
V:=LinearAlgebra[RandomVector](6,generator=10..20);
if nops(convert(V,set))=6 then break fi;
od:
V;

or

restart;
with(combinat):
randomize():
convert(randperm(randcomb([$ 10..20], 6)), Vector);

@vv  Thanks for this elaboration. It is clear that these inequalities  (a>2  and  so on) were found by hand.

Compare with solution in Mathematica:

@Carl Love  I started from the desired result for OP and came to the following interpretation: a set is given, each element of which is the product of two factors. One of these factors is fixed (in the question these are  a, b, c ). Find the set of remaining factors.

@Carl Love OP probably just misused the term "common factor". I have shown how to get the desired result in another obvious interpretation, which OP obviously had in mind.

@nm  I wrote "In each specific example ..."

@acer  You wrote  "And how about requesting their spacing evenly by arclength rather than by value of the independent parameter? "

See the post  https://www.mapleprimes.com/posts/213452-Uniform-Point-Plot  with a procedure for this.