Solveeit Logo
Login

Question

Question: Find the value of \( {\left( {1 - \omega + {\omega ^2}} \right)^5} + {\left( {1 - {\omega ^2} + \ome...

Find the value of (1−ω+ω2)5+(1−ω2+ω)5{\left( {1 - \omega + {\omega ^2}} \right)^5} + {\left( {1 - {\omega ^2} + \omega } \right)^5} , if ω\omega and ω2{\omega ^2} are complex cube roots of unity.

Explanation

Solution

First we have to define the terms we need to solve the problem.
Given that ω\omega and ω2{\omega ^2} are complex cube roots of unity.
Therefore,
ω3=1{\omega ^3} = 1
Use the identity,
1+ω+ω2=01 + \omega + {\omega ^2} = 0
Therefore,
1+ω=−ω21 + \omega = - {\omega ^2}
1+ω2=−ω1 + {\omega ^2} = - \omega
Formula Used:
ω3=1{\omega ^3} = 1
1+ω+ω2=01 + \omega + {\omega ^2} = 0

Complete step by step answer:
We need to find the value of (1−ω+ω2)5+(1−ω2+ω)5{\left( {1 - \omega + {\omega ^2}} \right)^5} + {\left( {1 - {\omega ^2} + \omega } \right)^5}
To find the required value let us solve the above equation,
(1−ω+ω2)5+(1−ω2+ω)5{\left( {1 - \omega + {\omega ^2}} \right)^5} + {\left( {1 - {\omega ^2} + \omega } \right)^5}
Let us rearrange the terms in the bracket,
=(1+ω2−ω)5+(1+ω−ω2)5= {\left( {1 + {\omega ^2} - \omega } \right)^5} + {\left( {1 + \omega - {\omega ^2}} \right)^5}
On substituting value of 1+ω2=−ω1 + {\omega ^2} = - \omega in first bracket we get,
=(−ω−ω)5+(1+ω−ω2)5= {\left( { - \omega - \omega } \right)^5} + {\left( {1 + \omega - {\omega ^2}} \right)^5}
On substituting value of 1+ω=−ω21 + \omega = - {\omega ^2} in second bracket we get,
=(−ω−ω)5+(−ω2−ω2)5= {\left( { - \omega - \omega } \right)^5} + {\left( { - {\omega ^2} - {\omega ^2}} \right)^5}
On performing the addition in first bracket we get,
=(−2ω)5+(−ω2−ω2)5= {\left( { - 2\omega } \right)^5} + {\left( { - {\omega ^2} - {\omega ^2}} \right)^5}
On performing the addition in second bracket we get,
=(−2ω)5+(−2ω2)5= {\left( { - 2\omega } \right)^5} + {\left( { - 2{\omega ^2}} \right)^5}
On performing the power operations of the first bracket we get,
=−32ω5+(−2ω2)5= - 32{\omega ^5} + {\left( { - 2{\omega ^2}} \right)^5}
On performing the power operations of the second bracket we get,
=−32ω5−32ω10= - 32{\omega ^5} - 32{\omega ^{10}}
On taking −32ω5 - 32{\omega ^5} from the both brackets we get,
=−32ω5(1+ω2)= - 32{\omega ^5}(1 + {\omega ^2})
On substituting value of 1+ω2=−ω1 + {\omega ^2} = - \omega in the bracket we get,
=−32ω5×(−ω)= - 32{\omega ^5} \times ( - \omega )
On performing the multiplication of both terms we get,
=32ω6= 32{\omega ^6}
On rearranging the terms we get,
=32×(ω3)2= 32 \times {({\omega ^3})^2}
On substituting value of ω3=1{\omega ^3} = 1 in the bracket we get,
=32×(1)2= 32 \times {(1)^2}
On performing the multiplication of both terms we get,
=32= 32
This is the required value.
Therefore,
(1−ω+ω2)5+(1−ω2+ω)5=32{\left( {1 - \omega + {\omega ^2}} \right)^5} + {\left( {1 - {\omega ^2} + \omega } \right)^5} = 32

Note: Any number when raised to power 33 or multiplied itself thrice, if given an answer as 11 , then such number is called the cube root of unity. Complex cube root is usually denoted by ω\omega . Complex cube roots of unity are imaginary cube roots of unity. One imaginary cube root of unity is the square of another imaginary cube root of unity i.e. one imaginary cube root of unity is ω\omega and other is ω2{\omega ^2} .