Question

If $1,\omega $ and ${\omega ^2}$ are the cube roots of unity, then \[(1 - \omega + {\omega ^2})(1 - {\omega ^2} + {\omega ^4})(1 - {\omega ^4} + {\omega ^8})(1 - {\omega ^8} + {\omega ^{16}})...2n\] is
A. $2n$
B. ${2^{2n}}$
C. $1$
D. $ - {2^{2n}}$

Answer 1

Hint:To solve this question we should know the two rules of cubes roots of omega $(\omega )$. The first rule is that $1 + \omega + {\omega ^2} = 0$. By transferring omega $(\omega )$ to the right hand side, we can write it as $1 + {\omega ^2} = - \omega $.And the second rule is that the cube of omega is one i.e. ${\omega ^3} = 1$. We will use these two properties to solve the above question by breaking the terms.

Complete step by step answer:
Here we have \[(1 - \omega + {\omega ^2})(1 - {\omega ^2} + {\omega ^4})(1 - {\omega ^4} + {\omega ^8})(1 - {\omega ^8} + {\omega ^{16}})...2n\] .
Let us take the first term
$(1 - \omega + {\omega ^2})$
From the above property we can write
$1 + {\omega ^2} = - \omega $
By putting this back in the expression we have :
$( - \omega - \omega )$
We will now take the second term i.e.
\[(1 - {\omega ^2} + {\omega ^4})\] .
We can write
${\omega ^4} = {\omega ^3} \times \omega $

And we know that
${\omega ^3} = 1$
So by putting this we get
${\omega ^4} = \omega \times 1$
We can put this in the expression and it gives us:
$(1 - {\omega ^2} + \omega )$
Now we take the third term of the given expression in the question :
\[(1 - {\omega ^4} + {\omega ^8})\] .
Here we can write
${\omega ^8} = {\omega ^6} \times {\omega ^2}$
Further we can write it as
${\omega ^3} \times {\omega ^3} \times {\omega ^2}$

It gives us the value
${\omega ^8} = 1 \times 1 \times \omega = \omega $
And from the above solution we can write
${\omega ^4} = \omega $
By putting all the values in the expression we can write;
$(1 - \omega + {\omega ^2})$
Let us now take the fourth term:
\[(1 - {\omega ^8} + {\omega ^{16}})\] .
Again we can write
${\omega ^8} = \omega $
And, also the value of
${\omega ^{16}} = \omega $
By putting this back in the expression we can write :
$(1 - \omega + {\omega ^2})$
Now we will write all the above terms back together in the original form i.e.
$( - \omega - \omega )(1 + \omega - {\omega ^2})(1 - \omega + {\omega ^2})(1 + \omega - {\omega ^2})..2n$

We will now add and simplify the terms, We can write
$1 + \omega = - {\omega ^2}$ from the rule of cube of omega
So we can write them now and multiply the first two terms together and the other two terms together:
$[ - 2\omega \times - 2{\omega ^2}][ - 2\omega \times - 2{\omega ^2}]...n$ terms
(because we make all the pairs of two terms, so it will have n terms in total)
Now on simplifying we have:
$[{2^2} \times {\omega ^3}][{2^2} \times {\omega ^3}]...n$ times and we know that ${\omega ^3} = 1$, so our new terms are
${2^2} \times {2^2} \times {2^2}...n$ times.
So if we multiply ${2^2}$ by n times, it gives us ${2^{2n}}$.

Hence the correct option is B.

Note:We should note that in the above solution we can write
 ${\omega ^{16}} = {\omega ^{15}} \times \omega $
And we can write ${\omega ^{15}}$ as
${\omega ^3} \times {\omega ^3} \times {\omega ^3} \times {\omega ^3} \times {\omega ^3}$.
So by putting this in the expression we can write
${\omega ^{16}} = {\omega ^3} \times {\omega ^3} \times {\omega ^3} \times {\omega ^3} \times {\omega ^3} \times \omega $
And we know the value of
${\omega ^3} = 1$
So it gives us value
${\omega ^{16}} = 1 \times \omega = \omega $