Question

If $1,\omega ,{\omega }^2$ are the cube roots of unity then $(3+3{\omega }^2+5\omega {)}^6-(2+6{\omega }^2+2\omega {)}^3$ is equal to

• A. $32$
• B. $64$
• C. $0$
• D. $1$

Answer 1

Answer: (C)

$0$

$\left(3\right(1+{\omega }^2)+5\omega {)}^6-(2(1+\omega )+6{\omega }^2{)}^3$
$=\left(3\right(-\omega )+5\omega {)}^6-(2(-{\omega }^2)+6\omega {)}^2{)}^3$
$=(2\omega {)}^6-(4{\omega }^2{)}^3$
$=2^6-4^3$
$=64-64=0$