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 $1+w+w^2=0$ ?

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Answer 1 · 2017-12-28T23:47:48

See below.

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

$x^2+x+1=0$

Using quadratic formula:

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-1+-sqrt((1)^2-4(1)(1)))/(2(1))=>x=(-1+sqrt(-3))/2$

$x = (-1-sqrt(-3))/2$

All roots:

$x=1 , x= (-1+sqrt(-3))/2, x= (-1-sqrt(-3))/2$

If $w = (-1+sqrt(-3))/2$

$w^2=((-1+sqrt(-3))/2)^2=(-1-sqrt(-3))/2$

$1+w+w^2=0$

$1+(-1+sqrt(-3))/2+((-1+sqrt(-3))/2)^2=0$