If w is one of the imaginary roots of the equation x3 = 1,
then the quartic (1 – w + w2)(1 + w – w 2) is equal to:
- 4
- w2
- 2
- –w
- 1
w3 = 1, so w3 – 1 = 0.
Now factor,
Since w is imaginary, w – 1 can’t be 0, so w2 + w + 1 = 0, or w2 = –w – 1.*
We want (1 – w + w2)(1 + w – w2). Substituting *, we have
= (1 – w + (-w – 1))(1 + w – (-w – 1))
= (1 – w – w -1)(1 + w + w + 1)
= (-2w)(2w + 2)
= (-2w)2(w + 1)
= -4w(w + 1)
= 4w(-w – 1)
= 4w(w2), using * again
= 4w3, but recall w3 = 1
= 4(1) = 4
= (1 – w – w -1)(1 + w + w + 1)
= (-2w)(2w + 2)
= (-2w)2(w + 1)
= -4w(w + 1)
= 4w(-w – 1)
= 4w(w2), using * again
= 4w3, but recall w3 = 1
= 4(1) = 4
The answer is (a).