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

Mar 12, 2018

The answer is -1-5i

#### Explanation:

First we factorize powers of i.So ${i}^{6}$ would become -1. $3 {i}^{3}$ would become -3i and ${i}^{2}$ would become -1.By putting values
$- 1 \left(2 i - \left(- 1\right) + 3 i\right)$. Now $- 1 \left(2 i + 3 i + 1\right)$ would become $- 1 \left(5 i + 1\right)$ and then $- 5 i - 1$
I HOPE IT HELPS:)

Mar 12, 2018

See details below

#### Explanation:

Expand expresion.

${i}^{6} \left(2 i - {i}^{2} - 3 {i}^{3}\right) = 2 {i}^{7} - {i}^{8} - 3 {i}^{9}$

But we know that:

${i}^{1} = i$
${i}^{2} = - 1$ (by definition)
i^3=i^2·i=-i
i^4=i^2·i^2=(-1)·(-1)=1
i^5=i^4·i=1·i=i and starts again
i^6=i^5·i=i·i=i^2=-1
i^7=i^6·i=-i

for this cliclic result, we can write:

${i}^{6} \left(2 i - {i}^{2} - 3 {i}^{3}\right) = 2 {i}^{7} - {i}^{8} - 3 {i}^{9} = - 2 i - 1 - 3 i = - 1 - 5 i$