Question

How do you find all the zeros of `f(x)=x^3-1`?

Answer 1

`f(x)` has zeros `1`, `omega = -1/2+sqrt(3)/2i` and `bar(omega)= -1/2-sqrt(3)/2i`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We use this below with `a=(x+1/2)` and `b=sqrt(3)/2i`.

`f(x) = x^3-1`

Note that `f(1) = 1-1 = 0` so `x=1` is a zero and `(x-1)` a factor...

`x^3-1`

`= (x-1)(x^2+x+1)`

`= (x-1)((x+1/2)^2-1/4+1)`

`= (x-1)((x+1/2)^2+3/4)`

`= (x-1)((x+1/2)^2-(sqrt(3)/2i)^2)`

`= (x-1)(x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)`

Hence the other two zeros are:

`x = -1/2+-sqrt(3)/2i`

One of these is often denoted by the Greek letter `omega`

`omega = -1/2+sqrt(3)/2i`

This is called the primitive Complex cube root of `1`.

The other is `bar(omega) = omega^2 = -1/2-sqrt(3)/2i`