Question

$1,\omega ,{\omega }^2$ are the cube roots of unity, then the value of $(1+w{)}^3-(1+{\omega }^2{)}^3$ is-

• A. $2\omega$
• B. $2$
• C. $-2$
• D. $0$

Answer 1

Answer: (D)

$0$

$(1+\omega {)}^3-(1+{\omega }^2{)}^3=(1+\omega -1-{\omega }^2)\left[\right(1+\omega {)}^2+(1+\omega )(1+{\omega }^2)+(1+{\omega }^2{)}^2]=(\omega -{\omega }^2)[1+{\omega }^2+2\omega +1+{\omega }^2+\omega +{\omega }^3+1+{\omega }^4+2{\omega }^2]=(\omega -{\omega }^2)[4+4\omega +4{\omega }^2]=(\omega -{\omega }^2)\times 0=0$