Question
If $$\omega \left( { \ne 1} \right)$$ is a cube root of unity and $${\left( {1 + \omega } \right)^7} = A + B\omega $$ then $$A$$ and $$B$$ are respectively
A.
$$0, 1$$
B.
$$1, 1$$
C.
$$1, 0$$
D.
$$- 1, 1$$
Answer :
$$1, 1$$
Solution :
$$\eqalign{
& {\left( {1 + \omega } \right)^7} = A + B\omega \cr
& \Rightarrow \,\,{\left( { - {\omega ^2}} \right)^7} = A + B\omega \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because 1 + \omega + {\omega ^2} = 0} \right) \cr
& \Rightarrow \,\, - {\omega ^{14}} = A + B\omega \cr
& \Rightarrow \,\, - {\omega ^2} = A + B\omega \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because {\omega ^3} = 1} \right) \cr
& \Rightarrow \,\,1 + \omega = A + B\omega \cr
& \Rightarrow \,\,A = 1,B = 1 \cr} $$