Question

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

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

Answer 1

Answer: (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