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If the roots of the equation${x^n} - 1 = 0$are$1,\alpha ,\beta ,\gamma ,.....$, show that$\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \gamma } \right)..... = n$

Last updated date: 22nd Jul 2024
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Given that: - roots of the equation${x^n} - 1 = 0$are$1,\alpha ,\beta ,\gamma ,.....$
$\therefore {x^n} - 1 = \left( {x - 1} \right)\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)..... \\ \Rightarrow \dfrac{{{x^n} - 1}}{{x - 1}} = \left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)..... \\$
Take $x \to 1$
We know that$\left[ {\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^n} - 1}}{{x - 1}} = n} \right]$ (using L-Hospitals rule)
$\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^n} - 1}}{{x - 1}} = \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \gamma } \right)..... \\ \Rightarrow n = \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \gamma } \right)..... \\ \therefore \left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \gamma } \right)..... = n \\$
Note: - L-Hospitals rule has been used here as the limit was in$\dfrac{0}{0}$ form. L-Hospitals rule can also be used for$\dfrac{\infty }{\infty }$ form. In this case the limiting value is differentiated and then the limit problem is preceded.