Question

If $\omega (\ne 1)$ be a cube root of unity and $(1+\omega {)}^7=A+B\omega$, where $A,B$ are real numbers, then the value of $A,B$ is :

• A. $0,1$
• B. $1,1$
• C. $1,0$
• D. $-1,1$

Answer 1

The correct option is B $1,1$

We know,
$1+\omega +{\omega }^2=0$ [sum of cube roots of unity $=0$]
$\Rightarrow 1+\omega =-{\omega }^2$
$\Rightarrow (1+\omega {)}^7=(-1{)}^7{\omega }^{14}=(-1{)}^7{\omega }^2$
$=-{\omega }^2$ [again $-{\omega }^2=1+\omega$]
$=1+\omega$
$A+B\omega$
$\Rightarrow A=1,B=1$