Painevg

2022-01-17

How do you simplify ${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)$?

intacte87

Solution:
Expand expression
${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)=2{i}^{7}-{i}^{8}-3{i}^{9}$
But we are conscious of:
${i}^{1}=i$
${i}^{2}=-1$ (by definition)
${i}^{3}={i}^{2}\cdot i=-i$
${i}^{4}={i}^{2}\cdot {i}^{2}=\left(-1\right)\cdot \left(-1\right)=1$
${i}^{5}={i}^{4}\cdot i=1\cdot i=i$ and starts again
${i}^{6}={i}^{5}\cdot i=i\cdot i={i}^{2}=-1$
${i}^{7}={i}^{6}\cdot i=-i$
To get this click-through result, type:
${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)=2{i}^{7}-{i}^{8}-3{i}^{9}=-2i-1-3i=-1-5i$

Jillian Edgerton

We factorize powers of "i" first. So ${i}^{6}$ would become $01.3{i}^{3}$ would become -3i and ${i}^{2}$ would become -1.By putting values
$-1\left(2i-\left(-1\right)+3i\right)$. Now $-1\left(2i+3i+1\right)$ would become $-1\left(5i+1\right)$ and then
$-5i-1$

alenahelenash

${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)=2{i}^{7}-{i}^{8}-3{i}^{9}$ ${i}^{1}=i$ ${i}^{2}=-1$ ${i}^{3}={i}^{2}\cdot i=-i$ ${i}^{4}={i}^{2}\cdot {i}^{2}=\left(-1\right)\cdot \left(-1\right)=1$ ${i}^{5}={i}^{4}\cdot i=1\cdot i=i$ ${i}^{6}={i}^{5}\cdot i=i\cdot i={i}^{2}=-1$ ${i}^{7}={i}^{6}\cdot i=-i$ ${i}^{6}\left(2i-{i}^{2}-3{i}^{3}\right)=2{i}^{7}-{i}^{8}-3{i}^{9}=-2i-1-3i=-1-5i$