Question

If $\omega (\ne 1)$ is a cube root of unity and $(1+\omega {)}^7=A+B\omega$, then $A$ and $B$, are repectively

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

Answer 1

The correct option is B 1, 1

Since $w$ is the cube root of unity.
$\therefore w^3=1\&1+w+w^2=0$
${(1+w)}^7=A+Bw\Rightarrow {(-w^2)}^7=A+Bw\Rightarrow {-\left(w^3\right)}^4{w}^2=A+Bw\Rightarrow -w^2=A+Bw\Rightarrow 1+w=A+Bw\therefore A=1\&B=1$
Ans: B