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

1 Answer
Aug 8, 2016

Answer:

#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#