Question

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`, show that if w is a complex root of this equation, the other complex root is `w^2` and `1+w+w^2=0`?

Answer 1

See below.

Explanation:

`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`