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Question: If \(1,\omega ,\;{\omega ^2},...............{\omega ^{n - 1}}\) are \(n,{n^{th}}\) roots of unity, t...

If 1,ω,  ω2,...............ωn−11,\omega ,\;{\omega ^2},...............{\omega ^{n - 1}} are n,nthn,{n^{th}} roots of unity, then the value of (9−ω)(9−ω2)(9−ω3)...............(9−ωn−1)\left( {9 - \omega } \right)\left( {9 - {\omega ^2}} \right)\left( {9 - {\omega ^3}} \right)...............\left( {9 - {\omega ^{n - 1}}} \right) will be
A. 9n+18\dfrac{{{9^n} + 1}}{8}
B. 9n−1{9^n} - 1
C. 9n−18\dfrac{{{9^n} - 1}}{8}
D. 9n+1{9^n} + 1

Explanation

Solution

In this question, we will proceed by taking x=(1)1nx = {\left( 1 \right)^{\dfrac{1}{n}}} and then raising on both sides by nn. Then convert this into an equation and apply binomial expansion. Further substitute x=9x = 9 to get the required answer.

Complete step-by-step answer:
Here we have to find the value of given expression (9−ω)(9−ω2)(9−ω3)...............(9−ωn−1)\left( {9 - \omega } \right)\left( {9 - {\omega ^2}} \right)\left( {9 - {\omega ^3}} \right)...............\left( {9 - {\omega ^{n - 1}}} \right)
Let’s say x=(1)1nx = {\left( 1 \right)^{\dfrac{1}{n}}}
And hence on taking power nn on both sides, we have
⇒xn=(11n)n ⇒xn=1 ⇒xn−1=0  \Rightarrow {x^n} = {\left( {{1^{\dfrac{1}{n}}}} \right)^n} \\\ \Rightarrow {x^n} = 1 \\\ \Rightarrow {x^n} - 1 = 0 \\\
We know that xn−1=(x−1)(x−ω)(x−ω2)...............(x−ωn−1){x^n} - 1 = \left( {x - 1} \right)\left( {x - \omega } \right)\left( {x - {\omega ^2}} \right)...............\left( {x - {\omega ^{n - 1}}} \right)
Dividing with x−1x - 1 on both sides, we have
⇒xn−1x−1=(x−ω)(x−ω2)...............(x−ωn−1)\Rightarrow \dfrac{{{x^n} - 1}}{{x - 1}} = \left( {x - \omega } \right)\left( {x - {\omega ^2}} \right)...............\left( {x - {\omega ^{n - 1}}} \right)
Put x=9x = 9, then we have
⇒9n−19−1=(9−ω)(9−ω2)..................(9−ωn−1) ⇒9n−18=(9−ω)(9−ω2)..................(9−ωn−1)  \Rightarrow \dfrac{{{9^n} - 1}}{{9 - 1}} = \left( {9 - \omega } \right)\left( {9 - {\omega ^2}} \right)..................\left( {9 - {\omega ^{n - 1}}} \right) \\\ \Rightarrow \dfrac{{{9^n} - 1}}{8} = \left( {9 - \omega } \right)\left( {9 - {\omega ^2}} \right)..................\left( {9 - {\omega ^{n - 1}}} \right) \\\
And hence the value of (9−ω).(9−ω2).(9−ω3)...............(9−ωn−1)\left( {9 - \omega } \right).\left( {9 - {\omega ^2}} \right).\left( {9 - {\omega ^3}} \right)...............\left( {9 - {\omega ^{n - 1}}} \right) is equals to
⇒9n−18\Rightarrow \dfrac{{{9^n} - 1}}{8}
Thus, the correct option is C. 9n−18\dfrac{{{9^n} - 1}}{8}

So, the correct answer is “Option C”.

Note: Here we have used the binomial expansion xn−1=(x−1)(x−ω)(x−ω2)...............(x−ωn−1){x^n} - 1 = \left( {x - 1} \right)\left( {x - \omega } \right)\left( {x - {\omega ^2}} \right)...............\left( {x - {\omega ^{n - 1}}} \right). Always remember that 1+ω+ω2=01 + \omega + {\omega ^2} = 0 and ω3=1{\omega ^3} = 1.