Solveeit Logo

Question

Question: The value of \(\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \rig...

The value of (x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8).....\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right).....till 2n2n factors.
A. (x1)2n{{\left( x-1 \right)}^{2n}}
B. 2(x1)2n+12{{\left( x-1 \right)}^{2n+1}}
C. (x1)2n1{{\left( x-1 \right)}^{2n-1}}
D. (x1)2n+1{{\left( x-1 \right)}^{2n+1}}

Explanation

Solution

We can observe that the terms ω,ω2\omega ,{{\omega }^{2}} are in the problem. So we have the relation between the cube roots of 11 which are 1,ω,ω21,\omega ,{{\omega }^{2}} as 1+ω+ω2=01+\omega +{{\omega }^{2}}=0. From this relation we will write the value of ω+ω2=1\omega +{{\omega }^{2}}=-1. We know that ω3=1{{\omega }^{3}}=1, ω4=ω{{\omega }^{4}}=\omega , ω5=ω2{{\omega }^{5}}={{\omega }^{2}}, ω6=ω3=1{{\omega }^{6}}={{\omega }^{3}}=1 and the sequence goes on. So, from the known values we will simplify the given equation (x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8).....\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right)..... to find the required value.

Complete step by step solution:
Given that, (x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8).....\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right).....
We have the value ω+ω2=1\omega +{{\omega }^{2}}=-1. Substituting the all the above values in the given equation, then we will have
(x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8).....=(x1)(x+ω2+ω4)(x+ω4+ω8).....\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right).....=\left( x-1 \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right).....
Now the values of ω4=ω{{\omega }^{4}}=\omega , ω8=(ω4)2=ω2{{\omega }^{8}}={{\left( {{\omega }^{4}} \right)}^{2}}={{\omega }^{2}}. Substituting these values in the above equation, then we will have
(x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8).....=(x1)(x+ω2+ω)(x+ω+ω2).....\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right).....=\left( x-1 \right)\left( x+{{\omega }^{2}}+\omega \right)\left( x+\omega +{{\omega }^{2}} \right).....
Rearranging the terms in an order, then we will get
(x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8).....=(x1)(x+ω+ω2)(x+ω+ω2).....\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right).....=\left( x-1 \right)\left( x+\omega +{{\omega }^{2}} \right)\left( x+\omega +{{\omega }^{2}} \right).....
Again, we have the value ω+ω2=1\omega +{{\omega }^{2}}=-1, substituting this value in above equation, then we will have
(x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8).....=(x1)(x1)(x1).....\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right).....=\left( x-1 \right)\left( x-1 \right)\left( x-1 \right).....
Here we have the value of (x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8).....\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right)..... but we need to calculate the value of (x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8).....\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right)..... till 2n2nterms, so
(x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8)..... till 2n terms=(x1)(x1)(x1)..... 2n times\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right).....\text{ till 2n terms}=\left( x-1 \right)\left( x-1 \right)\left( x-1 \right).....\text{ 2n times}
We know that xn=x.x.x.x.... till n terms{{x}^{n}}=x.x.x.x....\text{ till n terms}, hence
(x+ω+ω2)(x+ω2+ω4)(x+ω4+ω8)..... till 2n terms=(x1)2n\left( x+\omega +{{\omega }^{2}} \right)\left( x+{{\omega }^{2}}+{{\omega }^{4}} \right)\left( x+{{\omega }^{4}}+{{\omega }^{8}} \right).....\text{ till 2n terms}={{\left( x-1 \right)}^{2n}}
\therefore option A is the correct answer.

Note: There are similar problems like (1ω+ω2)(1ω2+ω4)(1ω4+ω8)..... till 2n factors.\left( 1-\omega +{{\omega }^{2}} \right)\left( 1-{{\omega }^{2}}+{{\omega }^{4}} \right)\left( 1-{{\omega }^{4}}+{{\omega }^{8}} \right).....\text{ till 2n factors}\text{.} For this kind of problem also we are going to use the same procedure, the same formula to get the answer. Some students might not realize that the question is related to cube roots of unity, i.e 1,ω,ω21,\omega,{{\omega }^{2}}. They may try to expand the whole thing using algebraic identities and then get stuck after a while. So, it is important to be able to relate the question with the topic and then answer.