Question

Show that (1+w-w^2)^6 = 64

Answer 1

Answer:

If w is one of the complex cube roots of unity, how can we show that 1+w equals 1? If w is a complex cube root of unity, how can we solve (2 + 2w - w2) 3 - (1-w+w2) 3? ... If 1, w, w² are cube roots of unity, then what is the value of (2-w) (2-w²) (2- w¹⁰) (2-w¹¹)

Answer 2

Answer:

1+W+W2=0

1+W=--W2

[-W2-W2]6

[-2W2]6

64W12

W12=1

64 1

64

Step-by-step explanation: