Find all roots of x^3-1, show that if w is a complex root of this equation, the other complex root is w^2 and 1+w+w^2=0?

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Find all roots of x^3-1, show that if w is a complex root of this equation, the other complex root is w^2 and 1+w+w^2=0?

Find all roots of $x^{3} - 1$ $x^3-1$ , show that if w is a complex root of this equation, the other complex root is $w^{2}$ $w^2$ and $1 + w + w^{2} = 0$ $1+w+w^2=0$ ?

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Answer 1 · 2017-12-28T23:47:48

See below.

Explanation:

$x^{3} - 1 = (x - 1) (x^{2} + x + 1)$ $x^3-1=(x-1)(x^2+x+1)$ ( difference of 2 cubes )

$x^{2} + x + 1 = 0$ $x^2+x+1=0$

Using quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{- 1 \pm \sqrt{{(1)}^{2} - 4 (1) (1)}}{2 (1)} \Rightarrow x = \frac{- 1 + \sqrt{- 3}}{2}$ $x=(-1+-sqrt((1)^2-4(1)(1)))/(2(1))=>x=(-1+sqrt(-3))/2$

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

All roots:

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

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

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

$1 + w + w^{2} = 0$ $1+w+w^2=0$

$1 + \frac{- 1 + \sqrt{- 3}}{2} + {(\frac{- 1 + \sqrt{- 3}}{2})}^{2} = 0$ $1+(-1+sqrt(-3))/2+((-1+sqrt(-3))/2)^2=0$

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Find all roots of x^3-1, show that if w is a complex root of this equation, the other complex root is w^2 and 1+w+w^2=0?

Find all roots of $x^{3} - 1$ $x^3-1$ , show that if w is a complex root of this equation, the other complex root is $w^{2}$ $w^2$ and $1 + w + w^{2} = 0$ $1+w+w^2=0$ ?

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Explanation:

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Explanation:

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Find all roots of $x^{3} - 1$ $x^3-1$ , show that if w is a complex root of this equation, the other complex root is $w^{2}$ $w^2$ and $1 + w + w^{2} = 0$ $1+w+w^2=0$ ?