Can it use Rationalizing the denominator on combination of 5th roots in the denominator

Question

I'm looking for simplify of this expression: $$ \frac{1}{\sqrt[5]{2} + \sqrt[5]{3} + \sqrt[5]{5}} = ? $$

I've tried to use the different of 5th power on them. But too complicated.

Answer 1

For simplicity let $a=\sqrt[5]{2}$, $b=\sqrt[5]{3}$ and $c=\sqrt[5]{5}$. Asking to rationalize the denominator of $(a+b+c)^{-1}$ is equivalent to asking what $(a+b+c)^{-1}$ looks like in the $\mathbb{Q}$-vector space $\mathbb{Q}(a,b,c)$.

Unfortunately, $\mathbb{Q}(a,b,c)$ is 125 dimensional over $\mathbb{Q}$. This means rationalizing the denominator amounts to solving a system of 125 linear equations with 125 variables.

I gave the computation to SageMath and this was the result:

$$\left(-\frac{22}{905} a b^{4} + \frac{13}{905} a^{2} b^{3} + \frac{17}{905} a^{3} b^{2} - \frac{53}{1810} a^{4} b + \frac{56}{905}\right) c^{4} + \left(\frac{9}{905} a^{2} b^{4} - \frac{6}{181} a^{3} b^{3} + \frac{19}{1810} a^{4} b^{2} - \frac{3}{905} b + \frac{2}{181} a\right) c^{3} + \left(\frac{21}{905} a^{3} b^{4} + \frac{41}{1810} a^{4} b^{3} - \frac{16}{905} b^{2} - \frac{7}{905} a b - \frac{37}{905} a^{2}\right) c^{2} + \left(-\frac{83}{1810} a^{4} b^{4} - \frac{5}{181} b^{3} + \frac{23}{905} a b^{2} + \frac{44}{905} a^{2} b - \frac{26}{905} a^{3}\right) c + \frac{108}{905} b^{4} + \frac{2}{905} a b^{3} - \frac{67}{905} a^{2} b^{2} - \frac{18}{905} a^{3} b + \frac{301}{1810} a^{4}$$