x³ - 3 = 0

    x³ = 3

So the zeros are the 3 cube roots of 3

3 = 3[cos(2pn) + i·sin(2pn)]

To get the three cube roots of 3, we use DeMoivre's theorem
using the power of 

  

∛3[cos + i·sin]

To get the first cube root, we let n=0

∛3[cos + i·sin] = ∛3[cos(0) + i·sin(0)] = ∛3[1 + 0i] = ∛3

To get the second cube root of 3, we let n=1

∛3[cos + i·sin] = ∛3[cos + i·sin] = ∛3[ + i·] =

 

and we can write 



So the second cube root of 3 is 

To get the third cube root of 3, we let n=2

∛3[cos + i·sin] = ∛3[cos + i·sin]

and since cos = cos, and sin = -sin,
 
the third cube root is just the conjugate of the 
second cube root, or



So the zeros of x³ - 3 are:

∛3, , and 

Edwin