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

##### 1 Answer

Aug 8, 2016

#### Answer:

#### Explanation:

The difference of squares identity can be written:

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

We use this below with

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

Note that

#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 = -1/2+sqrt(3)/2i#

This is called the primitive Complex cube root of

The other is