10) Arrange the three numbers $2 \sqrt[6]{2}, \sqrt{5}, \sqrt[3]{11}$ in ascending order.
(A) $\sqrt[3]{11}<\sqrt{5}<2 \sqrt[6]{2}$
(B) $\sqrt[3]{11}<2 \sqrt[6]{2}<\sqrt{5}$
(E) $2 \sqrt[6]{2}<\sqrt{5}<\sqrt[3]{11}$
(D) $\sqrt{5}<2 \sqrt[6]{2}<\sqrt[3]{11}$
10. The answer is $(\mathbf{A})$.

Recall that if $a>b>0$ and $n>0$ then $a^{n}>b^{n}$. Observe that by choosing $n=6$, which is the least common multiple of the indices of the roots (6, 2, 3 respectively), the comparison of the sixth powers of the given numbers reduces to a comparison of integers.

We therefore have:

$$
(2 \sqrt[6]{2})^{6}=2^{6} \cdot 2=128 . \quad(\sqrt{5})^{6}=5^{3}=125, \quad(\sqrt[3]{11})^{6}=11^{2}=121
$$

Since $121<125<128$ we have that $\sqrt[3]{11}<\sqrt{5}<2 \sqrt[6]{2}$.