If $$\omega$$ is the cube root of unity then $$(1 - \omega + \omega^2)^6 = $$
  • A
    $$0$$
  • A
    $$0$$
  • B
    $$1$$
  • B
    $$1$$
  • C
    $$64$$
  • C
    $$64$$
  • D
    $$32$$
  • D
    $$32$$

Solution

Given $$\omega$$ being cube root of unity.
Then,
$$(1 - \omega + \omega^2)^6  $$
$$=( - 2\omega )^6  $$ [Since $$1+\omega^2=-\omega$$]
$$=64$$ [Since $$\omega^3=1\Rightarrow \omega^6=1$$]