Question

Answer in the form (a+ib)
(i-1)^3​

Answer 1

We have to write (i-1)³ im form of (a+ib)

We know that :-

$\sf {(a - b)}^{3} = ( {a}^{2} + {b}^{2} - 2ab)(a - b)$

Therefore now :-

$\sf \implies {(i - 1)}^{3} = ( {i}^{2} + {1}^{2} - 2i)(i - 1)$

Where i = √(-1)

• i² = -1

$\sf \implies {(i - 1)}^{3} = ( - 1 + {1} - 2i)(i - 1)$

$\sf \implies {(i - 1)}^{3} = ( 0 - 2i)(i - 1)$

$\sf \implies {(i - 1)}^{3} = ( - 2i)(i - 1)$

$\sf \implies {(i - 1)}^{3} = - 2 {i}^{2} + 2 i$

$\sf \implies {(i - 1)}^{3} = - 2 ( - 1) + 2 i$

$\sf \implies {(i - 1)}^{3} = 2 + 2 i$

Answer :

$\bf 2 + 2 i$

Answer 2

Step-by-step explanation:

write (i-1)³ im form of (a+ib)

We know that :-

{(a - b)}^{3} = ( {a}^{2} + {b}^{2} - 2ab

(a- b)

Therefore:-

{(i - 1)}^{3} = ( {i}^{2} + {1}^{2} - 2i)(i-1)

⟹ (i−1) 3 =(i 2 +1 2 −2i)(i−1)

Where i = √(-1)

i² = -1

{(i - 1)}^{3} = ( - 1 + {1} - 2i)(i - 1)

⟹(i−1) 3 =(−1+1−2i)(i−1)

{(i - 1)}^{3} = ( 0 - 2i)(i - 1)

⟹(i−1) 3=(0−2i)(i−1)

{(i - 1)}^{3} = - 2 {i}^{2} + 2

⟹(i−1) 3=−2i2+2i

{(i - 1)}^{3} = 2 + 2 i

⟹(i−1) 3 =2+2i

Correct Answer :2 + 2 i2+2i