Question

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Answer 1

Answer:

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Answer 2

$\huge\mathfrak\red{Answer}$

${x}^{3} - 1 = 0 \\ {x}^{3} - {(1)}^{3} = 0 \\ (x - 1)( {x}^{2} + 1 + x) = 0 \\ \\ from \: here \: we \: get \\ x = 1 \\ \\ {x}^{2} + 1 + x = 0 \\ a = 1 \\ b = 1 \\ c = 1 \\ \\ x = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \: or \: \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = \frac{ - 1 + \sqrt{{(1)}^{2} - 4(1)(1)} }{2(1)} \\ \\ x = \frac{ - 1 + \sqrt{ - 3} }{2} \: or \: \frac{ - 1 - \sqrt{ - 3} }{2} \\ \\ x = \frac{ - 1 + i \sqrt{3} }{2} \: or \: \frac{ - 1 - i \sqrt{3} }{2} \\ \\ let \: w = \frac{ - 1 + i \sqrt{3} }{2} \\ squaring \: on \: both \: side \\ \\ {w}^{2} = \frac{ {( - 1 + i\sqrt{3}) }^{2} }{4} = \frac{ - 1 - i \sqrt{3} }{2} \\ \\ 1 + w + {w}^{2} \\ 1 + \frac{ - 1 + i \sqrt{3} }{2} + ( \frac{ - 1 - i \sqrt{3} }{2} ) \\ \\ \\ 1 + ( \frac{ - 1 + i \sqrt{3} - 1 - i \sqrt{3} }{2} \\ \\ 1 - 1 = 0$