Mathematica never assumes that it can choose the branch of a multi-valued function to match your imagination. It always uses the principal branch, or at least tries to (for definite integration, it sometimes gets lost in the complex plane). If you want to give it that choice, you must state your problem in terms that allow it. Perhaps like this:

Solve[{
  (I Sqrt[3] + 1) cr2/(2 crax) == -2, 
  cr2^3 == 2, 
  crax^3 == -2 a - 1},
   {a, cr2, crax}]

yielding

{{a -> -(5/8), cr2 -> -(-2)^(1/3), crax -> 1/4 ((-2)^(1/3) + I (-2)^(1/3) Sqrt[3])}, 
 {a -> -(5/8), cr2 -> 2^(1/3), crax -> 1/4 (-2^(1/3) - I 2^(1/3) Sqrt[3])}, 
 {a -> -(5/8), cr2 -> (-1)^(2/3) 2^(1/3), 
     crax -> 1/4 (-(-1)^(2/3) 2^(1/3) + (-1)^(1/6) 2^(1/3) Sqrt[3])}}

So, it even finds three choices of branches to get your result.

It isn't really the machinery of Solve that's in question.

f == 0 /. a -> -5/8
(* False *)

-5/8 isn't a root of your equation. Mathematica consistently treats x^(1/3) and CubeRoot[x] as different things, defined differently. If Solve were to violate that, it would represent a bug.

This is not a bug. 5/8 is simply not a solution to the given equation (others have pointed that out). Solve is working exactly as designed in this example.

To clarify, radicals give principal branches of roots. Parametrized radicals will of course only be able to do that when numeric values are substituted for the parameters.

As for Solve, when encountering a parametrized radical it creates a new variable and a new equation. For example, the expression (-1-2 a)^(1/3) would become something along the lines radicalVariable[(-1-2 a)^(1/3)] with the defining equation radicalVariable[(-1-2 a)^(1/3)]^3==(-1-2 a). The actual internals can be see below.

f = (I Sqrt[3] 2^(1/3) + 2^(1/3))/(2 (-2 a - 1)^(1/3));
In[3]:= Internal`MakePolynomial[f]

(* Out[3]= {{(Solve`ParmVar[2^(1/3)] + 
     Solve`ParmVar[I] Solve`ParmVar[2^(1/3)] Solve`ParmVar[Sqrt[
       3]]) Solve`RecipVar[1/
    Solve`RadVar[(-1 - 2 Solve`SolvVar[a])^(1/3)]], 
  1 + Solve`ParmVar[I]^2 == 0, -2 + Solve`ParmVar[2^(1/3)]^3 == 
   0, -3 + Solve`ParmVar[Sqrt[3]]^2 == 0, 
  1 + Solve`RadVar[(-1 - 2 Solve`SolvVar[a])^(1/3)]^3 + 
    2 Solve`SolvVar[a] == 
   0, -1 + Solve`RadVar[(-1 - 2 Solve`SolvVar[a])^(
      1/3)] Solve`RecipVar[1/
      Solve`RadVar[(-1 - 2 Solve`SolvVar[a])^(1/3)]] == 
   0}, {Solve`RecipVar[1/
   Solve`RadVar[(-1 - 2 Solve`SolvVar[a])^(1/3)]], 
  Solve`RadVar[(-1 - 2 Solve`SolvVar[a])^(1/3)], Solve`SolvVar[a], 
  Solve`ParmVar[I], Solve`ParmVar[Sqrt[3]], Solve`ParmVar[2^(1/3)]}} *)

This is not a bug.

Your rewrite using CubeRoot changes the meaning of the expression, because

(-2)^(1/3) == CubeRoot[-2]
(* False *)

or if you like

{(-2)^(1/3), CubeRoot[-2]} // N
{0.629961 + 1.09112 I, -1.25992}

In general, x^(1/3) refers to the principal branch of the complex cube root. Use CubeRoot or Surd for real roots.

Hi Hans,

This is a nice work around. But, this makes me believe that Solve has a deeper problem than I thought.

After simplifying to

(-2)^(1/3)/(-1-2 a)^(1/3)

Solve still cannot solve it unless you explicitly tell Solve that those 1/3 powers are cube roots! This means that Solve cannot properly recognize branching points if powers are given as rational fractions. I think the reason it worked with CubeRoot is because branching points are hardcoded in the function. But for any arbitrary rational fraction, Solve may fail again, which makes me worry.

I don't know how Wolfram programmed Solve, but I would expect Solve to try out the work around you suggested internally and give me the result.

The problem I posted here is actually a part of another problem I encountered with DSolve. DSolve correctly found the 3 roots of the general solution of a differential equation I gave it to solve, but when I asked Mathematica to solve the differential equation with a boundary condition, 2 out 3 solutions failed. I then manually applied the B.C. to the general solution Mathematica found with Solve, but I failed (as it is the topic of this thread). However, when I found a solution by hand, I realized that Solve has some issues and perhaps DSolve suffers from this as well. I reserve the problem I had with DSolve to another thread, so that we can focus on the problem Solve has.

In my opinion, this is a bug. To me, (-2)^(1/3)/(-1-2 a)^(1/3)==-2 is a very simplified expression and Solve cannot solve it without an unnecessary replacement of 1/3 powers with CubeRoot.

Someone with better mathematics knowledge than I have will need to answer. To show my ignorance level I would argue that x==2 means that x==2 only. Analogous to your cubing of both sides x^2==4 is a different equation that results in x==2 and x==-2.

Are you sure that a=-5/8 is a solution?

f = (I Sqrt[3] 2^(1/3) + 2^(1/3))/(2 (-2 a - 1)^(1/3));
f /. a -> -5/8 // FullSimplify
(* 1 + I Sqrt[3] *)

Note that

Solve[f == 2, a]
(* {{a -> -(3/8)}} *)