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Question: If \(\omega ( \ne 1)\) is a cube root of unity and \({(1 + {\omega ^2})^n} = {(1 + {\omega ^4})^n}\)...

If ω(≠1)\omega ( \ne 1) is a cube root of unity and (1+ω2)n=(1+ω4)n{(1 + {\omega ^2})^n} = {(1 + {\omega ^4})^n}, then the least positive value of n is:
A. 2
B. 3
C. 5
D. 6

Explanation

Solution

We know that ω3=1{\omega ^3} = 1 , So ω3−1=0{\omega ^3} - 1 = 0 , so we can use the expansion of a3−b3{a^3} - {b^3} for expanding ω3−1{\omega ^3} - 1. After that, we will compare the LHS with RHS to get a new equation, we can find some more relevant equations. After which we will substitute all values in the equation given in the question to get to the final answer.

Complete step by step solution:
Let us assume the cube root of unity as ω\omega , mathematically
ω3=1{\omega ^3} = 1= … (1)
Shifting RHS term(s) to LHS, we get
⇒ω3−1=0\Rightarrow {\omega ^3} - 1 = 0
This can also be written as,
⇒ω3−13=0\Rightarrow {\omega ^3} - {1^3} = 0 … (2)
We know that,
a3−b3=(a−b)(a2+ab+b2){a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}) … (3)
From (3) and (2), we can say that
⇒ω3−13=(ω−1)(ω2+ω+1)\Rightarrow {\omega ^3} - {1^3} = (\omega - 1)({\omega ^2} + \omega + 1) … (4)
Now, since ω≠1\omega \ne 1 (given), we get
⇒(ω2+ω+1)=0\Rightarrow ({\omega ^2} + \omega + 1) = 0 … (5)
From (5), we can say that
⇒1+ω2=−ω\Rightarrow 1 + {\omega ^2} = - \omega … (6)
And,
⇒1+ω=−ω2\Rightarrow 1 + \omega = - {\omega ^2} … (7)
Also,
⇒1+ω4=1+ω3ω\Rightarrow 1 + {\omega ^4} = 1 + {\omega ^3}\omega
Since, ω3=1{\omega ^3} = 1 , we get
⇒1+ω4=1+ω\Rightarrow 1 + {\omega ^4} = 1 + \omega … (8)

Using (7) in (8), we get
⇒1+ω4=−ω2\Rightarrow 1 + {\omega ^4} = - {\omega ^2} … (9)
We are given that
⇒(1+ω2)n=(1+ω4)n\Rightarrow {(1 + {\omega ^2})^n} = {(1 + {\omega ^4})^n}
Swapping LHS by RHS, we get
⇒(1+ω4)n=(1+ω2)n\Rightarrow {(1 + {\omega ^4})^n} = {(1 + {\omega ^2})^n}
Divide the above equation by (1+ω4)n{(1 + {\omega ^4})^n} , we get
⇒(1+ω4)n(1+ω2)n=1\Rightarrow \dfrac{{{{(1 + {\omega ^4})}^n}}}{{{{(1 + {\omega ^2})}^n}}} = 1 … (10)
Put values of (6) and (9) in (10), we get
⇒(−ω2)n(−ω)n=1\Rightarrow \dfrac{{{{( - {\omega ^2})}^n}}}{{{{( - \omega )}^n}}} = 1
After simplification, we will get
⇒ωn=1\Rightarrow {\omega ^n} = 1
Hence, n=3,6,9,12,...n = 3,6,9,12,...

And the least positive value of n is 3.
Hence the correct option is B.

Note:
The cube roots of unity can be defined as the numbers which when raised to the power of 3 gives the result as 1, That is why we assumed ω3=1{\omega ^3} = 1 . Some formulas can come in handy to solve these types of questions.
(1) (ω2+ω+1)=0({\omega ^2} + \omega + 1) = 0
(2) ω3−13=(ω−1)(ω2+ω+1){\omega ^3} - {1^3} = (\omega - 1)({\omega ^2} + \omega + 1)
(3) ω3=1{\omega ^3} = 1
While solving these types of questions these formulas would surely help.