Solve $16x^{-3}=-2$ 
Solve $16x^{-3}=-2$. 

My working:
\begin{align}
16x^{-3}&=-2\\
\frac{1}{16x^{3}}&=-2\\
\frac{16x^3}{16x^3}&=-32x^3\\
1&=-32x^{3}\\
-32x^{3}&=1\\
-32x&=\sqrt[3]{1}\\
-32x&=1\\
x&=\frac{-1}{32}
\end{align}
Is this right? What have I done wrong? 

$$16x^{-3} = -2$$
$$\frac{16}{x^3} = -2$$
$$-8 = x^3$$
$$x = -2$$
