Your first example: If you drop the third argument then Solve will return all 3 roots. The real root is, obviously, the first one.

Solve[y^3+x==0,y]
{{y->-x^(1/3)},{y->(-1)^(1/3) x^(1/3)},{y->-(-1)^(2/3) x^(1/3)}}

Your second example: Take the real root and use Simplify on it.

realRoot=Solve[x-(8+1/y)^3==0,y]//First
Simplify[realRoot,Assumptions->Element[x,PositiveReals]]

To select the real roots (if any) you could try this function:

realRoots[r_List]:=Select[r,FreeQ[#,(-1)^_]&&FreeQ[#,I]&&FreeQ[#,Complex]&]

The Power[] expression x^(1/3) is not real-valued for negative x. Are you sure that's what you want?

In other words, you cannot represent the real solutions as a parametrized Power[] expression, except as a Piecewise[] expression.

Principal-value real radicals are represented by Surd in Mathematica.

The Solve/Reduce complex of solvers implement real roots via Root[]. In general, solutions cannot be expressed in terms of Surd[]. For instance, the solutions to something as seemingly simple as 8 x^3 == 1 + 6 x, which are three real roots, cannot be expressed in terms of Surd[]. This is, I suppose, why Wolfram has adopted Root[] for expressing polynomial roots.

So, yeah, it's a problem. I'm not sure how to help. Or if I know how to help.

Of course, the solutions to certain forms of equations can be expressed in terms of Surd[]. Do all your equations have one of the forms u^n + a == 0 or u^(-n) + a == 0, where u and a are expressions in terms of y and x and n is an integer?

No. Not all. I'm searching for a more general solution.