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

If the roots of the equationxn−1=0{x^n} - 1 = 0are1,α,β,γ,.....1,\alpha ,\beta ,\gamma ,....., show that(1−α)(1−β)(1−γ).....=n\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {1 - \gamma } \right)..... = n

Explanation

Solution

Hint: - This problem can be solved by using limits and L-Hospital’s rule.
Given that: - roots of the equationxn−1=0{x^n} - 1 = 0are1,α,β,γ,.....1,\alpha ,\beta ,\gamma ,.....

∴xn−1=(x−1)(x−α)(x−β)(x−γ)..... ⇒xn−1x−1=(x−α)(x−β)(x−γ).....  \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→1x \to 1
We know that[lim⁡x→1xn−1x−1=n]\left[ {\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^n} - 1}}{{x - 1}} = n} \right] (using L-Hospitals rule)
⇒lim⁡x→1xn−1x−1=(1−α)(1−β)(1−γ)..... ⇒n=(1−α)(1−β)(1−γ)..... ∴(1−α)(1−β)(1−γ).....=n  \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 \\\
Hence the equation is proved.

Note: - L-Hospitals rule has been used here as the limit was in00\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.