The 14th Thailand Mathematical Olympiad

Let $r=\sqrt[3]{p}+\sqrt[3]{p^5}$. Then, $$r^3=(\sqrt[3]{p}+\sqrt[3]{p^5}{)}^3=p+p^5+3p^2(\sqrt[3]{p}+\sqrt[3]{p^5})=p+p^5+3p^{2}r.$$ Hence $r$ is a root of the polynomial $x^3-3p^{2}x-p^5-p$. Assume to the contrary that $r$ is rational. By the rational root theorem we have $r$ is an integer. From $r^3=p+p^5+3p^{2}r$ we get $p\mid r$. Hence $p^3\mid r^3-3p^{2}r-p^5$, implying $p^3\mid p$ a contradiction. Thus, $r$ is irrational.

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Alternative solution.

Assume to the contrary that $r=\sqrt[3]{p}+\sqrt[3]{p^5}$ is rational. Thus $$\frac{r+p^3}{1+pr}=\frac{\sqrt[3]{p}+\sqrt[3]{p^5}+\sqrt[3]{p^9}}{1+\sqrt[3]{p^4}+\sqrt[3]{p^8}}=\sqrt[3]{p}$$ is rational. This is clearly false, thus $r$ is irrational.