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

Hint: The unity cube root is symbolized as $\sqrt[3]{1}$, and it has three roots. The three roots of the cube root of unity are \[1,{\text{ }}\omega ,{\text{ }}{\omega ^2}\], which when multiplied together yields the answer unity. One of the roots of the cube root of unity is a real root, whereas the other two are imaginary roots.

Formula Used:
${\omega ^3} = 1$
$1 + \omega + {\omega ^2} = 0$

Complete step by step solution:
The Cube root of Unity is any number that gives the answer one when raised to the power of three or multiplied by itself three times. The cube root of any number is the number that, when raised to the power of three, yields the number whose cube root is to be found.
Let, \[1,{\text{ }}\omega ,{\text{ }}{\omega ^2}\] are the cube roots of unity
$ \Rightarrow {\omega ^3} = 1,1 + \omega + {\omega ^2} = 0$ Or $1 + \omega = - {\omega ^2} - - - - - \left( 1 \right)$
Given that,
${\left( {1 + \omega } \right)^7} = A + B\omega $
From equation (1),
${\left( { - {\omega ^2}} \right)^7} = A + B\omega $
$ - {\left( {{\omega ^3}} \right)^4}{\omega ^2} = A + B\omega $
$ - {\omega ^2} = A + B\omega $
Again, using equation (1)
$1 + \omega = A + B\omega $
Comparing both the sides,
$A = 1,B = 1$

Option ‘B’ is correct

Note: The key concept involved in solving this problem is the good knowledge of cube root of unity. Students must remember that in cube root of unity the product of imaginary roots is one and the sum of the cube roots is equal to zero where imaginary roots are \[\omega ,{\text{ }}{\omega ^2}\].