# How do you multiply (-2i)^3?

Jul 31, 2016

$8 i$

#### Explanation:

First, ruse the rule that ${\left(a b\right)}^{3} = {a}^{3} \cdot {b}^{3}$. That is:

${\left(- 2 i\right)}^{3} = {\left(- 2\right)}^{3} \cdot {i}^{3}$

We can see easily that ${\left(- 2\right)}^{3} = - 2 \left(- 2\right) \left(- 2\right) = - 8$.

To determine the value of ${i}^{3}$, first write it as ${i}^{2} \cdot i$. Since $i = \sqrt{- 1}$, we see that ${i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1$.

Thus, ${i}^{3} = {i}^{2} \cdot i = - 1 \cdot i = - i$.

Putting this all together, we see that:

${\left(- 2 i\right)}^{3} = {\left(- 2\right)}^{3} \cdot {i}^{3} = \left(- 8\right) \cdot \left(- i\right) = 8 i$