Question

The value of $\left( {1 - \omega - {\omega ^2}} \right)\left( {1 - {\omega ^2} + {\omega ^4}} \right)\left( {1 - {\omega ^4} + {\omega ^8}} \right)..............$ to 2n factors is?
$\left( a \right){2^n}$
$\left( b \right){4^n}$
$\left( c \right)4n$
$\left( d \right)2n$

Answer 1

In this particular question se the concept that here $\omega$ is a cube root of unity so it satisfies the conditions such as, ${\omega ^3} = 1$ and $1 + \omega + {\omega ^2} = 0$. So use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation:
$\left( {1 - \omega - {\omega ^2}} \right)\left( {1 - {\omega ^2} + {\omega ^4}} \right)\left( {1 - {\omega ^4} + {\omega ^8}} \right)..............$ to 2n factors
Now as we know that $\omega$ is a cube root of unity so it satisfies some of the conditions such as,
${\omega ^3} = 1$................ (1)
And
$1 + \omega + {\omega ^2} = 0$................... (2)
Now first simplify the given equation we have,
$\Rightarrow \left( {\left( {1 + {\omega ^2}} \right) - \omega } \right)\left( {1 - {\omega ^2} + {\omega ^3}.\omega } \right)\left( {1 - {\omega ^3}.\omega + {{\left( {{\omega ^3}} \right)}^2}{\omega ^2}} \right)..............$ to 2n factors
Now from equation (1) and (2) we have,
$\Rightarrow \left( { - \omega - \omega } \right)\left( {1 - {\omega ^2} + \left( 1 \right)\omega } \right)\left( {1 - \left( 1 \right).\omega + {{\left( 1 \right)}^2}{\omega ^2}} \right)..............$ to 2n factors
Now simplify the above equation we have,
$\Rightarrow \left( { - \omega - \omega } \right)\left( {\left( {1 + \omega } \right) - {\omega ^2}} \right)\left( {\left( {1 + {\omega ^2}} \right) - \omega } \right)..............$ to 2n factors
Now again from equation (2) we have,
$\Rightarrow \left( { - \omega - \omega } \right)\left( { - {\omega ^2} - {\omega ^2}} \right)\left( { - \omega - \omega } \right)..............$ to 2n factors
$\Rightarrow \left( { - 2\omega } \right)\left( { - 2{\omega ^2}} \right)\left( { - 2\omega } \right)..............$ to 2n factors
Now the above equation as 2n factors, so we can say that the above equation has n factors of $\left( { - 2\omega } \right)$ and n factors of $\left( { - 2{\omega ^2}} \right)$.
So the above equation is written as
$\Rightarrow \left( { - 2\omega } \right)\left( { - 2{\omega ^2}} \right)\left( { - 2\omega } \right)..............$ to 2n factors = ${\left( { - 2\omega } \right)^n}{\left( { - 2{\omega ^2}} \right)^n}$
Now simplify it we have,
$\Rightarrow {\left( { - 1} \right)^n}{2^n}{\omega ^n}{\left( { - 1} \right)^n}{2^n}{\left( {{\omega ^2}} \right)^n}$
Now the above equation is also written as
$\Rightarrow {\left( { - 1 \times - 1} \right)^n}{2^{n + n}}{\left( {\omega .{\omega ^2}} \right)^n}$
$\Rightarrow {\left( 1 \right)^n}{2^{2n}}{\left( {{\omega ^3}} \right)^n}$
Now as we know that any power of 1 is nothing but 1 and from equation 1 we have,
$\Rightarrow {2^{2n}}{\left( 1 \right)^n}$
$\Rightarrow {2^{2n}}$
$\Rightarrow {\left( {{2^2}} \right)^n}$
$\Rightarrow {\left( 4 \right)^n}$
So this is the required answer.

So, the correct answer is “Option b”.

Note: Whenever we face such types of questions the key concept we have to remember is the properties of the cube root of unity which is all stated above, so first simplify the above equation according to the cube root of unity properties as above simplified we will get the required answer.