Question

If $1,\omega $ and ${\omega ^2}$ are the cube roots of unity, then the value of $\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right)\left( {1 + {\omega ^4}} \right)\left( {1 + {\omega ^8}} \right)$ is equal to
A. $1$
B. $0$
C. ${\omega ^2}$
D. $\omega $

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: 
Given that,
\[1,{\text{ }}\omega ,{\text{ }}{\omega ^2}\] are the cube roots of unity
$ \Rightarrow {\omega ^3} = 1,1 + \omega + {\omega ^2} = 0$ Or $1 + {w^2} = - w$
Now, $\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right)\left( {1 + {\omega ^4}} \right)\left( {1 + {\omega ^8}} \right)$
$ = \left( {1 + \omega } \right)\left( { - \omega } \right)\left( {1 + \left( {{\omega ^3} \times \omega } \right)} \right)\left( {1 + \left( {{\omega ^3} \times {\omega ^3} \times {\omega ^2}} \right)} \right)$
$ = \left( {1 + \omega } \right)\left( { - \omega } \right)\left( {1 + \omega } \right)\left( {1 + {\omega ^2}} \right)$
$ = {\left( {1 + \omega } \right)^2}\left( { - \omega \left( {1 + {\omega ^2}} \right)} \right)$
$ = {\left( {1 + \omega } \right)^2}\left( { - \omega - {\omega ^3}} \right)$
$ = - {\left( {1 + \omega } \right)^3}$
$ = - {\left( { - {\omega ^2}} \right)^3}$
$ = {\omega ^6}$
$ = {\omega ^3} \times {\omega ^3}$
$ = 1$
Therefore, the correct option is (A)..

Note: The key concept involved in solving this problem is the good knowledge of the cube root of unity. Students must remember that in the 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 \[\omega ,{\text{ }}{\omega ^2}\].