Express ( 3 - 5i)³ in the form (a + bi )
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Answered by
2
We know that (a - b)³ = a³ - b³ - 3ab (a -b)
= ( 3 - 5i)³ = 3³ - (5i)³ -3 × 3 × 5i × ( 3 - 5i)
= 27 - 125i³ - 135i + 225i²
= ( 27 + 125i - 135i - 225)
= ( - 198 - 10i)
= ( 3 - 5i)³ = 3³ - (5i)³ -3 × 3 × 5i × ( 3 - 5i)
= 27 - 125i³ - 135i + 225i²
= ( 27 + 125i - 135i - 225)
= ( - 198 - 10i)
Answered by
6
(3 - 5i)³
use formula ,
(a - b)³ = a³ - 3a²b + 3ab² - b³
(3 - 5i)³ = (3)³ - 3(3)²(5i) + 3(3)(5i)²-(5i)³
= 27 -27×5i + 9×25(i²) -(125i³)
we know,
i² = -1
i³ = -i use it
= 27 -135i +225(-1) -125(-i)
= 27 - 225 -135i +125i
= -198 -10i
hence , ( -198) + (-10)i is in the form of (a + bi)
use formula ,
(a - b)³ = a³ - 3a²b + 3ab² - b³
(3 - 5i)³ = (3)³ - 3(3)²(5i) + 3(3)(5i)²-(5i)³
= 27 -27×5i + 9×25(i²) -(125i³)
we know,
i² = -1
i³ = -i use it
= 27 -135i +225(-1) -125(-i)
= 27 - 225 -135i +125i
= -198 -10i
hence , ( -198) + (-10)i is in the form of (a + bi)
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