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

Question

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?

Answer 1

You're doing two things wrong.

1. Interpreting $16x^{-3}$ as $\frac{1}{16x^3}$, it means $\frac{16}{x^3}$.

2. Only taking the cube root of $x^3$ in $-32x^3$.

Answer 2

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

Answer 3

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

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

$$-8 = x^3$$

$$x = -2$$

Answer 4

Using Maxima I obtain the following:

[x=1-sqrt(3)%i,x=sqrt(3)%i+1,x=-2]

Where %i = sqrt(-1)