How do you find three cube roots of $1$ $1$ ?

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Answer 1 · 2017-06-26T07:42:56

Three cube roots of $1$ $1$ are

$1$ $1$ , $\frac{- 1 - i \sqrt{3}}{2}$ $(-1-isqrt3)/2$ and $\frac{- 1 + i \sqrt{3}}{2}$ $(-1+isqrt3)/2$

Let $x = \sqrt[3]{1}$ $x=root(3)1$ , then $x^{3} = 1$ $x^3=1$ or

$x^{3} - 1 = 0$ $x^3-1=0$

or $(x - 1) (x^{2} + x + 1) = 0$ $(x-1)(x^2+x+1)=0$

Hence either $x = 1$ $x=1$

or $x^{2} + x + 1 = 0$ $x^2+x+1=0$ and using quadratic formula

$x = \frac{- 1 \pm \sqrt{1 - 4 \times 1 \times 1}}{2}$ $x=(-1+-sqrt(1-4xx1xx1))/2$

= $\frac{- 1 \pm \sqrt{- 3}}{2}$ $(-1+-sqrt(-3))/2$

= $\frac{- 1 \pm i \sqrt{3}}{2}$ $(-1+-isqrt3)/2$

Hence three cube roots of $1$ $1$ are

$1$ $1$ , $\frac{- 1 - i \sqrt{3}}{2}$ $(-1-isqrt3)/2$ and $\frac{- 1 + i \sqrt{3}}{2}$ $(-1+isqrt3)/2$

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