Write in descending order the numbers

$$
\sqrt[5]{3}, \quad 3^{-2 / 3}, \quad 3^{-2}, \quad\left(\frac{1}{3}\right)^{3} \text { and }\left(\frac{1}{3}\right)^{-1}
$$
Observe that $0<\frac{1}{c}<\frac{1}{b}<\frac{1}{a} \Longleftrightarrow 0<a<b<c$. Since $1<3^{2 / 3}<3^{2}<3^{3}$, it follows that

$$
\frac{1}{3^{3}}<\frac{1}{3^{2}}<\frac{1}{3^{2 / 3}}<1
$$

Also, $1<3<3^{5}$, so $1<\sqrt[5]{3}<3$. It remains to observe that $\frac{1}{3^{3}}=\left(\frac{1}{3}\right)^{3}, \frac{1}{3^{2}}=3^{-2}$, $\frac{1}{3^{2 / 3}}=3^{-2 / 3}$ and $3=\left(\frac{1}{3}\right)^{-1}$, to establish that

$$
\left(\frac{1}{3}\right)^{3}<3^{-2}<3^{-2 / 3}<\sqrt[5]{3}<\left(\frac{1}{3}\right)^{-1}
$$