If *w* is one of the imaginary roots of the equation *x*^{3} = 1,

then the quartic (1 – *w* + *w*^{2})(1 + *w* – w ^{2}) is equal to:

- 4
*w*^{2}- 2
- –
*w* - 1

*w*^{3} = 1, so *w*^{3} – 1 = 0.

Now factor, *w*^{3} – 1 = (*w* – 1)(*w*^{2} + *w* + 1).

Since *w* is imaginary, *w* – 1 can’t be 0, so *w*^{2} + *w* + 1 = 0, or *w*^{2} = –*w* – 1.*

We want (1 –

*w*+*w*^{2})(1 +*w*–*w*^{2}). Substituting *, we have
= (1 –

= (1 –

= (-2

= (-2

= -4

= 4

= 4

= 4

= 4(1) = 4

*w*+ (-*w*– 1))(1 +*w*– (-*w*– 1))= (1 –

*w*–*w*-1)(1 +*w*+*w*+ 1)= (-2

*w*)(2*w*+ 2)= (-2

*w*)2(*w*+ 1)= -4

*w*(*w*+ 1)= 4

*w*(-*w*– 1)= 4

*w*(*w*^{2}), using * again= 4

*w*^{3}, but recall*w*^{3}= 1= 4(1) = 4

The answer is (a).