Question

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

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

Answer 1

The correct option is B

$1,1$

Explanation for the correct option

We know that when $\omega$ is a cube root of unity then

${\omega }^3=1\text{and}1+\omega +{\omega }^2=0$ which can be rewritten as $1+\omega =-{\omega }^2$

As given that $A+B\omega$ is equal to ${\left(1+\omega \right)}^7$ then

$A+B\omega =\left(1+\omega \right){\left(1+\omega \right)}^6$

Since $1+\omega =-{\omega }^2$

$\begin{array}{rcl}A+B\omega & =& \left(1+\omega \right){\left(-{\omega }^2\right)}^6\\ & =& \left(1+\omega \right)\left({\omega }^{12}\right)\\ & =& \left(1+\omega \right){\left({\omega }^3\right)}^4\end{array}$

Also ${\omega }^3=1$ implies that

$\begin{array}{rcl}A+B\omega & =& 1+\omega \end{array}$

Therefore, the values of $A\text{and}B$ are $1\text{and}1$

Hence, the correct option is (B).