6.18. Determine whether the number is rational

$$
\alpha=\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}
$$
6.18. Answer: irrational. Let $\alpha_{1}=\sqrt[3]{4+\sqrt{15}}$ and $\alpha_{2}=$ $=\sqrt[3]{4-\sqrt{15}}$. Then $\alpha_{1}^{3}+\alpha_{2}^{3}=8$ and $\alpha_{1}^{3} \alpha_{2}^{3}=1$, so $\alpha_{1} \alpha_{2}=1$. Therefore, $\left(\alpha_{1}+\alpha_{2}\right)^{3}=\alpha_{1}^{3}+\alpha_{2}^{3}+3 \alpha_{1} \alpha_{2}\left(\alpha_{1}+\alpha_{2}\right)=8+3\left(\alpha_{1}+\alpha_{2}\right)$. Thus, $\alpha^{3}=8+3 \alpha$. Therefore, it remains to prove that the polynomial $x^{3}-3 x-8$ has no rational roots. According to problem 10.3, the rational roots of this polynomial are integers that are divisors of the number 8. Direct verification shows that the numbers $\pm 1, \pm 2, \pm 4, \pm 8$ are not roots of this polynomial.