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Question: Simplify the expression: \(\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-{{\omeg...

Simplify the expression:
(1−ω)(1−ω2)(1−ω4)(1−ω8)\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-{{\omega }^{4}} \right)\left( 1-{{\omega }^{8}} \right)

Explanation

Solution

In the above expression, ω\omega is the cube root of unity. And we know that a cube of ω\omega is equal to 1 which will look as follows: ω3=1{{\omega }^{3}}=1. Also, we know that sum of 1, ω\omega and square of ω\omega is equal to 0 and the mathematical expression for this addition will look as follows: 1+ω+ω2=01+\omega +{{\omega }^{2}}=0. Now, to simplify the above expression, we are going to write the powers of ω\omega in the form of ω3{{\omega }^{3}} so that we can write and then simplify using basic algebra.

Complete step-by-step solution:
The expression given in the above problem is as follows:
(1−ω)(1−ω2)(1−ω4)(1−ω8)\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-{{\omega }^{4}} \right)\left( 1-{{\omega }^{8}} \right)
In the above expression, ω\omega is the cube root of unity and the mathematical form of this cube root of unity is as follows:
ω=(1)13\omega ={{\left( 1 \right)}^{\dfrac{1}{3}}}
Cubing both the sides of the above equation we get,
ω3=1{{\omega }^{3}}=1
Also, there is a relation of the cube root of unity as follows:
1+ω+ω2=01+\omega +{{\omega }^{2}}=0
Now, we are going to rearrange the given expression by writing ω4=ω3(ω){{\omega }^{4}}={{\omega }^{3}}\left( \omega \right) and ω8=ω6(ω2){{\omega }^{8}}={{\omega }^{6}}\left( {{\omega }^{2}} \right) in the above expression and we get,
(1−ω)(1−ω2)(1−ω3(ω))(1−ω6(ω2))\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-{{\omega }^{3}}\left( \omega \right) \right)\left( 1-{{\omega }^{6}}\left( {{\omega }^{2}} \right) \right) ……… (1)
We have shown above that the cube of the cube root of unity is equal to 1.
ω3=1{{\omega }^{3}}=1
Now, taking square on both the sides of the above equation we get,
(ω3)2=(1)2 ⇒ω6=1 \begin{aligned} & {{\left( {{\omega }^{3}} \right)}^{2}}={{\left( 1 \right)}^{2}} \\\ & \Rightarrow {{\omega }^{6}}=1 \\\ \end{aligned}
Using the above relation in eq. (1) we get,
(1−ω)(1−ω2)(1−(1)(ω))(1−(1)(ω2)) =(1−ω)(1−ω2)(1−ω)(1−ω2) =(1−ω)2(1−ω2)2 \begin{aligned} & \left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-\left( 1 \right)\left( \omega \right) \right)\left( 1-\left( 1 \right)\left( {{\omega }^{2}} \right) \right) \\\ & =\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right) \\\ & ={{\left( 1-\omega \right)}^{2}}{{\left( 1-{{\omega }^{2}} \right)}^{2}} \\\ \end{aligned}
Rearranging the powers in the above expression we get,
((1−ω)(1−ω2))2{{\left( \left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right) \right)}^{2}}
Multiplying (1−ω)&(1−ω2)\left( 1-\omega \right)\And \left( 1-{{\omega }^{2}} \right) in the above expression and we get,
(1−ω2−ω+ω3)2 =(1−(ω2+ω)+ω3)2.......(3) \begin{aligned} & {{\left( 1-{{\omega }^{2}}-\omega +{{\omega }^{3}} \right)}^{2}} \\\ & ={{\left( 1-\left( {{\omega }^{2}}+\omega \right)+{{\omega }^{3}} \right)}^{2}}.......(3) \\\ \end{aligned}
In the above, we have shown that:
1+ω+ω2=01+\omega +{{\omega }^{2}}=0
Subtracting 1 on both the sides we get,
ω+ω2=−1\omega +{{\omega }^{2}}=-1
Using the above relation in eq. (3) we get,
(1−(−1)+1)2 =(1+1+1)2 =32=9 \begin{aligned} & {{\left( 1-\left( -1 \right)+1 \right)}^{2}} \\\ & ={{\left( 1+1+1 \right)}^{2}} \\\ & ={{3}^{2}}=9 \\\ \end{aligned}
From the above solution, we have solved the given expression to 9.
Hence, the simplification of the above expression is equal to 9.

Note: To solve the above problem, you must know the properties of the cube root of unity otherwise you cannot solve this problem so make sure you have a good understanding of this concept of the cube root of unity. Also, don’t make any calculation mistakes in the above problem.