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suppose $f(x)= x^3-2$ is the minimal polynomial for $\theta$. How can I find a minimal polynomial for $a = \frac{\theta^5}{5}$

My idea was to substitute $u = \sqrt[5]{5 \cdot x} $

and then plug it into $f(x)$ and reduce it to a minimal polynomial.

While $f(u)$ would have $a$ as a root, it may not be in $\mathbb{Q}[x]$.

So my question is, is there any method to find a polynomial or even better a minimal polynomial?

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    A "better" minimal polynomial? There is only one up to a multiple by a degree zero polynomial (=a number different from zero) – 2017-01-12
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    If $\theta^3=2$ then $\theta^5=2\theta^2$. – 2017-01-12
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    Sorry, maybe it is a punctuation error. I meant ..a polynomial, or even better, a minimal polynomial. – 2017-01-12

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So we have that

$$\;\theta\in\left\{\,\sqrt[3]2,\,\sqrt[3]2\,\omega\,,\,\,\sqrt[3]2\,\omega^2\,\right\}\;,\;\;\omega:=e^{2\pi i/3}\implies\frac{\theta^5}5\in\left\{\,\frac{2^{5/3}}5\,,\,\,\frac{2^{5/3}\omega^2}5\,,\,\,\frac{2^{5/3}\omega}5\,\right\}$$

and now:

$$r=\frac{\theta^5}5\implies125r^3=2^5=32\implies g(x):=x^3-\frac{32}{125}\in\Bbb Q[x]$$

makes the cut.

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    but it looks like I can use this method only, if I can separate $\theta$. what if $\theta$ is a root of $x^3-x-2$? – 2017-01-12
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    @user315567 Then things get way tougher... – 2017-01-12