#### Answer

negative

#### Work Step by Step

We know that $a^{\frac{m}{n}}=(\sqrt[n] a)^{m}$ (where m and n are positive integers greater than 1 with $\frac{m}{n}$ in simplest form and $\sqrt[n] a$ is a real number).
Therefore, $(-9)^{\frac{3}{2}}=(\sqrt -9)^{3}$.
However, $\sqrt -9$ is not a real number, as we cannot find some number $b$ such that $b^{2}=-9$.
In general, we cannot perform a radical function of a negative number when the index of the radical is even.