## The simplest way is to solve the equation in x
X:=fsolve(1/x^3-sin(12*x)=0, x=1.2..2, maxsols=10); # there are 3 solutions

     X := 1.266058342, 1.591679996, 1.818677859

Y := seq(1/x^3, x=X);

     Y := 0.4927638533, 0.2479891760, 0.1662389018

## As a system
s:={fsolve([y-sin(12*x)=0,x^3*y=1], {x,y}, {x=1.2..2})}; # first root

     s := {{x = 1.818677859, y = 0.1662389017}}

fsolve([y-sin(12*x)=0,x^3*y=1], {x,y}, {x=1.2..2}, avoid=s): # repeat if you want
s:=s union {%};

     s := {{x = 1.266058343, y = 0.4927638522}, {x = 1.818677859, y = 0.1662389017}}

N := 100:
for a to N do
for b from a to N do
for c from b to N do
  s:=[isolve(t^3-(a+b+c)*t^2+(a*b+a*c+b*c)*t-2*a*b*c)];
  if nops(s)>1 then print(rhs~(op~(s)), [a,b,c]) fi;
od od od:

You have a restart after with(LinearAlgebra).

So, P[0], P[1] are given and you need P[i] for i>1 in terms of L[i,j].
We may suppose P[0] = [0,0] (the origin) and P[1] = [a,0], a>0 (actually a=L[0,1]).
Only L[k,0], L[k,1] will be needed (k>1).

There are 2 solutions (symmetric wrt the x axis), so we may take the y-coordonate of P[k] be positive.

solve( [ x^2+y^2 = L[k,0]^2, (x-a)^2+y^2 = L[k,1]^2 ], {x,y}, explicit ):
select( u -> (sign(eval(y,u))=1), %):
P[k]=eval([x,y],%);

Use
G1:=RelabelVertices(G,[1,2,3,4,5,6]);

In Maple, a polynomial cannot have symbolic exponents (powers). For example, x^m - x^2 - 2 is not a polynomial, unless m is a (constant) nonnegative integer. This is normal; try for example to compute by hand the quotient and the remainder for the polynomials  x^m - x^2 - 2  and x^3 + 2*x + 3; such operations are essential in the Groebner package.

In conclusion your problem cannot be solved in a CAS (but you may solve it providing integer value(s) for m).

Maple needs a little help.

restart;
eq := x^2+floor(x)-10:
seq( solve({eval(eq, floor(x)=n), n <= x, x < n+1}), n=-5..5);

        {x = -sqrt(14)}, {x = 2*sqrt(2)}

Download LongestPath-sent2.mw

For such expressions involving principal branches, the solutions are difficult to find, and more important, they are much more difficult to check.

Consider the equation (equivalent to yours):
g:=2*ln(u) + ln(u^2+2);
solve( g - c = 0, u);       # 2 solutions     (1)
solve( g - ln(c) = 0, u);  # 4 solutions   (2)

If we are interested in real solutions (so, u>0, c real) then there is a unique solution.

But for complex u, c, do we have 2? 4? more? Of course (1) and (2) are contradictory (the 4 solutions given in (2) are distinct but not all are true -- for the principal branch of ln: just take a numeric c and check the solutions).
So, it's much more important to determine the true solutions rather than to eliminate the duplicates!

seq(expand(eval(test,_B1=v)), v=0..1); # 1/2, 1/2

Make the first multiplication inert.

restart;
expr:= A*B*C;
o1:= op(1,expr);
expr1:=o1%*(expr/o1);
nops(expr1);  # 2
value(expr1); # = expr

@Hnx 

diff  knows piecewise, so you can use it directly:

restart:
with(CurveFitting):
S := Spline([[0,0],[1,1],[2,4],[3,3]], v);
dS := diff(S,v);

floor(fsolve(1/(n+1)! = 0.0001, n)) + 1;

You can't use Dirac like that. Dirac is not a function; it is a distribution and must be handled with great care, and only if the user knows exactly its meaning. In Maple the use of the Dirac "function" is restricted to some contexts, e.g. as in the input and output of dsolve and in integral transforms (Fourier, Laplace).

The answer is no.
You will have to use the theory of Parametric LP.