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

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

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Answer 1 · 2018-03-12T10:24:39

The answer is -1-5i

Explanation:

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

Answer 2 · 2018-03-12T10:26:28

See details below

Explanation:

Expand expresion.

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

But we know that:

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

for this cliclic result, we can write:

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

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

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How do you simplify i^6(2i-i^2-3i^3) i6(2i−i2−3i3)i^6(2i-i^2-3i^3) ?

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