Find the complex cube root of 8

We have t find the complex cube root of 8

The cube roots of 8 are 2, 2ω and 2ω2 where ω=−12+√32i is the primitive Complex cube root of 1.

Solution

The cube roots of 8 plotted in the Complex plane on the circle of radius 2

They can be expressed as:

2(cos(0)+isin(0))=2

2(cos(2π3)+isin(2π3))=−1+√3i=2ω

2(cos(4π3)+isin(4π3))=−1−√3i=2ω2

One way of finding these cube roots of 8 is to find all of the roots of x3−8=0.

x3−8=(x−2)(x2+2x+4)

The quadratic factor can be solved using the quadratic formula:

x=−b±√b2−4ac2a

=−2±√22−(4×1×4)2⋅1

=−2±√−122

=−1±√3

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