Using [resultants][1] can greatly help. For example, knowing that $\sqrt[5]{2}$ is a zero of $P(x)=x^5-2$ and $1-\exp\frac{2\pi i}{5}$ is a zero of $Q(x)=(x-1)^5+1$, we conclude $\sqrt[5]{2}\cdot (1-\exp\frac{2\pi i}{5})$ is a zero of $\mathrm{Res}_y(P(y),y^5Q(\frac{x}{y}))$. In PARI/GP:

    ? polresultant(y^5-2,y^5*((x/y-1)^5+1),y)
    %1 = x^25 + 2500*x^15 + 50000*x^5
    ? factor(%)
    %2 = 
    [                       x 5]

    [x^20 + 2500*x^10 + 50000 1]

That is, we conclude that the minimal polynomial is $x^{20} + 2500x^{10} + 50000$. Numerical verification:

    ? subst(x^20 + 2500*x^10 + 50000, x, 2^(1/5)*(1-exp(2*Pi*I/5)) )
    %3 = 5.380530270144733809 E-34 - 2.104034275277802617 E-34*I

(the default precision is 38 decimal digits)



  [1]: https://en.wikipedia.org/wiki/Resultant#Number_theory