Question

If ${\rm{\omega }} \ne 1$ is a cube root of unity, and ${\left( {1 + {\rm{\omega }}} \right)^7} = {\rm{A}} + {\rm{B\omega }}$. Then (A, B) equals.
A) (0, 1)
B) (1, 1)
C) (1, 0)
D) (-1, 1)

Answer 1

Conditions for w to be cube root of unity are: ${{\rm{\omega }}^3} = 1$ and $1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0$. Using these conditions, form two equations and solve for values of A and B.

Complete step by step solution:
Since w is the cube root of unity,
${{\rm{\omega }}^3} = 1$ and $1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0$ (i)
Given, ${\left( {1 + {\rm{\omega }}} \right)^7} = {\rm{A}} + {\rm{B\omega }}$
$\Rightarrow {\left( { - {{\rm{\omega }}^2}} \right)^7} = {\rm{A}} + {\rm{B\omega }}$ $\left[ {{\rm{since}},{\rm{\;}}1 + {\rm{\omega }} = - {{\rm{\omega }}^2}} \right]$
$\Rightarrow - {\left( {{{\rm{\omega }}^3}} \right)^4} \times {{\rm{\omega }}^2} = {\rm{A}} + {\rm{B\omega }}$
$\Rightarrow - {\left( 1 \right)^4} \times {{\rm{\omega }}^2} = {\rm{A}} + {\rm{B\omega }}$ [Equation (i)]
$\Rightarrow - {{\rm{\omega }}^2} = {\rm{A}} + {\rm{B\omega }}$
$\Rightarrow 1 + {\rm{\omega }} = {\rm{A}} + {\rm{B\omega }}$ [Equation (ii)]
So, A = 1 and B =1

So, option (B) is correct.

Note:
Conditions of cube root of unity ${{\rm{\omega }}^3} = 1$ and $1 + {\rm{\omega }} + {{\rm{\omega }}^2} = 0$ are necessary to solve this type of questions. Substitute the values using the conditions to form equations.