Mathematica can solve the equation Solve[v'[r] == 0, r] analytically in terms of Root objects, as it is an algebraic equation of degree 6.

I would not call r = g^3/(2*r^2)... a "result", it is another equation with respect to r. I tried giving simple numerical values to the parameters, solving the two equations and comparing the results:

Block[{g = 1, M = 1},
 Solve[v'[r] == 0, r, Reals]]
Block[{g = 1, M = 1},
 \[Xi] = ((3 + 8*Pi*r^2)^3*(g^3 + r^3)^3)/(216*r^6) + 
   Sqrt[g^6 - (g^3*(3 + 8*Pi*r^2)^3*(g^3 + r^3)^3)/(108*r^6)] - g^3;
 NSolve[r == 
   g^3/(2*r^2) + 
    r/2 + (4/3)*(g^3 + r^3)*P*
     Pi + ((3 + 8*Pi*r^2)^2*(g^3 + r^3)^2)/(36*
       r^4*\[Xi]^(1/3)) + \[Xi]^(1/3),
  r, Reals]]