Question

show that (-1+i√3➗2)^3 is a real no. ​

Answer 1

Answer:

= 1 ( real number)

Step-by-step explanation:

did you mean -

$( \frac{ - 1 + i \sqrt{3} }{2} ) {}^{3}$

$= ( \frac{ - 1 + i \sqrt{3} }{2} ) {}^{3} \\ = \frac{1}{8} ({ - 1 + i \sqrt{3} }{} ) {}^{3}$

$\\ = \frac{1}{8} ({ i \sqrt{3} - 1 }{} ) {}^{3}$

expand using general formula of (a-b)³ :

$= \frac{1}{8} ( - i3 \sqrt{3} - 1 - 9 {i}^{2} + - i3 \sqrt{3} )$

= 1 ( real number)

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