Dear John,

I failed to understand your test. The way I defined f in the first post never meant to be equal to 0 for a=-5/8. There is no surprise that you got False.

Secondly, let's look at the correctly defined case

FullSimplify@f/.a->-5/8
(* 2(-1)^(1/3) *)

Even if you had used the original condition, you would still get False when you test it against -2 since Mathematica returns the principle root of 2(-1)^(1/3) which is 1.+1.73205I

2(-1)^(1/3)//N
(* 1.+1.73205*I *)

For a second, let's forget about how Mathematica evaluates an expression.

The expression 2(-1)^(1/3) has 3 roots. The roots are

2Exp[I Pi/3], 2Exp[I 5Pi/3], 2Exp[I Pi] 

2Exp[I Pi/3] is the principle root which Mathematica returns and its value is 1.+1.73205I. The real root is 2Exp[I Pi] which is returned by CubeRoot function and its value is -2 and finally the other root 2Exp[I 5Pi/2] whose value is 1.-1.73205I

The way I looked at Solve function initially was that I thought it was trying all these roots to find a, but it wasn't, as cleared out by Daniel.

If I am asked to check whether f==-2 for a=-5/8, I would consider all the roots. Not just the principle root which obviously doesn't satisfy the condition, but the root 2Exp[I Pi] does!

Back to Mathematica, you may consider the equations corresponding to each root as separate equations. Mathematica may choose to do that for any reason-- consistency is a very valid reason. Then you may say a=-5/8 does not satisfy the condition for my equation corresponding to the "principle root", which you are correct. But for another root, it satisfies.

I think you should now see the source of my initial objection (or say, confusion) which is related to the "machinery" of Solve. I'll get my head wrapped around the way Mathematica operates in the future.

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.

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)}} *)