Finding solution of expression with fractional power

edit

We do not know what is sol, but i suppose it is the following...

sage: var('r');
sage: eq = 3*(2.2 + (64/r)^(1/3)) == 4*(2.2 + (128/(r-1))^(1/4))
sage: sol = solve(eq, r)
sage: sol
[r^(1/3) == 60*(r - 1)^(1/4)/(40*8^(1/4) + 11*(r - 1)^(1/4))]

It is clear that we cannot ask for the numerical value of sol[0].

Instead, we could help sage deliver the needed value. We just substitute $$ r = R^3 $$ and rewrite the given equation in the form: $$ \frac 34\left(\frac{22}{10} +\frac 4R\right)-\frac {22}{10} = \left(\frac {128}{R^3-1}\right)^{1/4}\ . $$ Rising both sides to the fourth power leads to a polynomial equation, which we can solve. We obtain one value of $R$, introduced by taking the fourth power. But it leads to no solution in $r$.

Try a plot of the difference to see there is no solution.

If there is some typo, then fix it, find an interval where the difference function changes the sign, then use gp-solve. (Inside pari or inside sage.)

Ahem... solve suggests numerical approximate solutions :

sage: SN=solve(Eq, r, to_poly_solve=True) ; Sn
[r == (-0.08510234586942406 - 0.5483016107521035*I),
 r == (-0.08510234586942406 + 0.5483016107521035*I)]
sage: print([(Eq.lhs()-Eq.rhs()).subs(s).n() for s in SN])
[1.77635683940025e-15*I, -1.77635683940025e-15*I]

Not that far of the target... But indeed :

sage: print([(Eq.lhs()-Eq.rhs()).subs(s).is_zero() for s in SN])
[False, False]

Let's hunt for exact solutions, if any... Start by rewriting the equation, as judiciously suggested by dan_fulea :

Eq=3*((11/5)+(64/r)^(1/3))==4*(11/5+(128/(r-1))^(1/4)) ; print(Eq)

$$\frac{12}{r^{\frac{1}{3}}} + \frac{33}{5} = \frac{8 \cdot 8^{\frac{1}{4}}}{{\left(r - 1\right)}^{\frac{1}{4}}} + \frac{44}{5}$$

Work on this a bit :

E2=Eq.subs([r==R^3,(R^3)^(-1/3)==R^-1])-44/5 ; print(E2)

(One will note the awkward internal representation of powers in fractions...)

$$\frac{12}{R} - \frac{11}{5} = \frac{8 \cdot 8^{\frac{1}{4}}}{{\left(R^{3} - 1\right)}^{\frac{1}{4}}}$$

We eliminate the fractional powers by elevating both hands at the fourth power (thus potentially introducing spurious solutions...) :

E3=(E2^-4).factor()

$$\frac{625 \, R^{4}}{{\left(11 \, R - 60\right)}^{4}} = \frac{1}{32768} \, {\left(R^{2} + R + 1\right)} {\left(R - 1\right)}$$

This equation is utimately the equation of a rational fraction to zero. Its roots are therefore the roots of its denominator, provided that the denominator is not null in these points :

P=(E3.lhs()-E3.rhs()).factor().numerator()

$$-14641 \, R^{7} + 319440 \, R^{6} - 2613600 \, R^{5} + 29998641 \, R^{4} - 13279440 \, R^{3} + 2613600 \, R^{2} - 9504000 \, R + 12960000$$

A raincheck doesn't hurt :

sage: P.is_polynomial(R)
True

The roots of this polynomial are potential solutions of our original equations...

Sol=P.roots(x=R,ring=QQbar, multiplicities=False)

... provided that they do not nullify the denominator...

Sol=[u for u in Sol if not (E3.lhs()-E3.rhs())(R=u).is_zero()]

... and their cubes are indeed solution of the original equation :

sage: SolR=[u^3 for u in Sol if bool(Eq(r=u^3))] ; SolR
[-0.085102345869424? - 0.548301610752104?*I,
 -0.085102345869424? + 0.548301610752104?*I]

Those two solutions are "exact" in the same sense as QQbar is "exact" : there is a way to determinate them with as much as precision as needed, with a upper bound of the difference as small as desired, notwithstanding the absence of a closed-form radical expression...

HTH,

EDIT : This result is also found by Mathematica, as shown in the following atrocity :

sage: MS=mathematica.Solve(Eq,r);MS
{{r -> Root[-2176782336000000000000 + 6584000661504000000000*#1 - 
      10160322906612096000000*#1^2 + 21200873670270453024000*#1^3 - 
      24022442465994700088721*#1^4 - 15246001414346517837*#1^5 - 
      15216891738794163*#1^6 + 3138428376721*#1^7 & , 4, 0]}, 
 {r -> Root[-2176782336000000000000 + 6584000661504000000000*#1 - 
      10160322906612096000000*#1^2 + 21200873670270453024000*#1^3 - 
      24022442465994700088721*#1^4 - 15246001414346517837*#1^5 - 
      15216891738794163*#1^6 + 3138428376721*#1^7 & , 5, 0]}}
sage: MSR=union(SR(repr(MS[1][1][2][1]).replace("#1","x").replace("&","")).roots(ring=QQbar, multiplicities=False),SR(repr(MS[2][1][2][1]).replace("#1","x").replace("&","")).roots(ring=QQbar, multiplicities=False))
sage: [u for u in MSR if bool(Eq.subs(r=u))]
[-0.0851023458694239? + 0.5483016107521033?*I,
 -0.0851023458694239? - 0.5483016107521033?*I]