# How do you simplify i^3(2i^6-4i^21)?

Feb 9, 2016

${i}^{3} \left(2 {i}^{6} - 4 {i}^{21}\right) = 2 i - 4$

#### Explanation:

You should use the power laws

[1] $\text{ } {a}^{m} \cdot {a}^{n} = {a}^{m + n}$

[2] $\text{ } {\left({a}^{m}\right)}^{n} = {a}^{m \cdot n}$

Also, remember that

[3] $\text{ } {i}^{2} = - 1$

and with the help of [3], you can compute

[4] $\text{ } {i}^{4} = {i}^{2} \cdot {i}^{2} = \left(- 1\right) \cdot \left(- 1\right) = 1$

Thus, you can simplify as follows:

${i}^{3} \left(2 {i}^{6} - 4 {i}^{21}\right) = {i}^{3} \cdot 2 {i}^{6} - {i}^{3} \cdot 4 {i}^{21}$

$\stackrel{\text{[1] }}{=} 2 {i}^{3 + 6} - 4 {i}^{3 + 21}$

$= 2 {i}^{9} - 4 {i}^{24}$

Now, let's evaluate ${i}^{9}$ and ${i}^{24}$:

${i}^{9} = {i}^{4 + 4 + 1} = {i}^{4} \cdot {i}^{4} \cdot {i}^{1} \stackrel{\text{[4] }}{=} 1 \cdot 1 \cdot i = i$

${i}^{24} = {i}^{6 \cdot 4} = {i}^{6 \cdot 4} \stackrel{\text{[2] }}{=} {\left({i}^{4}\right)}^{6} = {1}^{6} = 1$

${i}^{3} \left(2 {i}^{6} - 4 {i}^{21}\right) = 2 {i}^{9} - 4 {i}^{24} = 2 i - 4$