Question

If $\omega \ne 1$ be a cube root of unity and ${\left(1+\omega \right)}^7=1+m\omega$, then the value of $l+m=$

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

Answer: (C)

$2$

$(1+\omega {)}^7=l+m\omega$

$1+\omega +{\omega }^2=0$

$1+\omega {)}^7=(1+\omega {)}^2(1+\omega {)}^2(1+\omega {)}^2(1+\omega )$

$=(1+{\omega }^2+2\omega )(1+{\omega }^2+2\omega (1+{\omega }^2+2\omega \left)\right(1+\omega$

$=(1+\omega +{\omega }^2+\omega )(1+{\omega }^2+\omega +\omega )(1+{\omega }^2+\omega +\omega )(1+\omega +{\omega }^2-{\omega }^2)$

$(0+\omega )(0+\omega )(0+\omega )(0-{\omega }^2)$ from 1 $1+\omega +{\omega }^2=0$

$=-{\omega }^5$

$=-\left({\omega }^3\right)\left({\omega }^2\right)$
$=-{\omega }^2$

$(1+\omega {)}^7=1+\omega$

$(1+\omega {)}^7=1+\omega =l+m\omega$

om comparing

$l=1$ and $m=1$

$\therefore l+m=1+1=2$