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if 1 omega omega 2 ldots omega n 1 are n th roots
Question:
If \( 1, \omega, \omega^2, \ldots, \omega^{n-1} \) are \( n \)th roots of unity, then the value of \( (9-\omega)(9-\omega^2)\cdots(9-\omega^{n-1}) \) is
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Always use $x^n - 1$ identity for roots of unity products.
MET - 2014
MET
Updated On:
Apr 23, 2026
$\frac{9^n + 1}{8}$
$9^n - 1$
$\frac{9^n - 1}{8}$
$9^n + 1$
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The Correct Option is
C
Solution and Explanation
Concept:
Product of $(x - \omega^k)$ identity
Step 1:
\[ (x-1)(x-\omega)\cdots(x-\omega^{n-1}) = x^n - 1 \]
Step 2:
Put $x = 9$: \[ (9-1)(9-\omega)\cdots(9-\omega^{n-1}) = 9^n - 1 \]
Step 3:
Divide by $(9-1)=8$: \[ (9-\omega)(9-\omega^2)\cdots = \frac{9^n - 1}{8} \]
Conclusion:
Answer = $\frac{9^n - 1}{8}$
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