Question

Simplify:
(root5 + 7 iota) (root 5 -7 oita)^3
plzz help me to solve it
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Answer 1

(√5 + 7i) (√5 - 7i)³

= {(√5 + 7i)(√5 - 7i)}(√5 - 7i)² .....(1)

we know, (a - b)(a + b) = a² - b²

so, (√5 - 7i)(√5 + 7i) = (√5)² - (7i)²

= 5 - 49(-1) [ as you know, i² = -1 ]

= 54

now, you can use (√5 + 7i)(√5 - 7i) = 54 in equation (1),

so, 54(√5 - 7i)²

using formula, (a - b)² = a² - 2ab + b²

= 54[(√5)² + (7i)² - 2√5(7i) ]

= 54[ 5 + 49(-1) - 14√5i ]

= 54 [ - 44 - 14√5 i]

= - 2376 - 756√5 i

hence, (√5 + 7i)(√5 - 7i)³ converts into -2376 - 756√5 i