Question

How do you simplify [(1+i)4(2−2i)3]?

Answer 1

Answer:

(1+i)⁴(2−2i)³ = 64(1 + i)

Step-by-step explanation:

(1+i)⁴(2−2i)³

= ((1 + i)²)²(2³)(1 -i)³

= ((1 + i)²)²(2³)(1 -i)²(1-i)

= 8 ( 1 + i² + 2i)² (1 + i² - 2i) ( 1 - i)

i² = -1

= 8( 1 - 1 + 2i)² (1 - 1 -2i) (1 - i)

= 8 (2i)²(-2i)( 1-i)

= 8 * 4i² (-2i)( 1-i)

= 32 (-1)(-2i) ( 1 - i)

= 64 i ( 1 - i)

= 64 i - 64i²

= 64i + 64

= 64 + 64 i

= 64(1 + i)

Answer 2

Answer:

$(1+i)^4(2-2i)^3=64(1+i)$

Step-by-step explanation:

Formula used:

$(a+b)(a-b)=a^2-b^2$

$i=\sqrt{-1}$

Given:

$(1+i)^4(2-2i)^3$

$=(1+i)(1+i)^3\:2^3\:(1-i)^3$

$=8(1+i)[(1+i)^3(1-i)^3]$

$=8(1+i)[(1+i)(1-i)]^3$

$=8(1+i)[1^2-i^2]^3$

$=8(1+i)[1+1]^3$

$=8(1+i)[2]^3$

$=8(1+i)[8]$

$=64(1+i)$