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? 

Obviously you could check that $x=-\frac1{32}$ doesn't solve $16x^{-3} = -2$.
Instead you can solve as follows
$$\begin{align*}
16 x^{-3} & = -2  & \text{reciprocal } a=b \Leftrightarrow \frac1a = \frac1b \\
\Leftrightarrow \frac1{16} x^3 & = -\frac12 & \text{multiply by } 16\\
\Leftrightarrow x^3 & = -8 & \text{take cube root}\\
\Leftrightarrow x & = \sqrt[3]{-8} = -2
\end{align*}$$
