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 $x^3 - 3p^2x - p^5 - p$. Assume to the contrary that $r$ is rational. By the rational root theorem we have $r$ is an integer. From $r^3 = p + p^5 + 3p^2r$ we get $p \mid r$. Hence $p^3 \mid r^3 - 3p^2r - p^5$, implying $p^3 \mid p$ a contradiction. Thus, $r$ is irrational.

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.