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Problem: The equation $x^n + px - q = 0$ has a double root. Show that $\left(\dfrac{p}{n}\right)^n + \left(\dfrac{q}{n-1}\right)^{n-1} = 0$.

My attempt: Let the double root be $\alpha$ and $P(x) = x^n + px - q$, then $P(\alpha) = 0$ and $P'(\alpha) = 0$. From $P'(\alpha) = 0$ we have $\alpha^{n-1} = -\dfrac{p}{n}$, substituting this into $P(\alpha) = 0$, we have $\alpha = \dfrac{qn}{p(n-1)}$. Then I tried subsituting this into $P(\alpha) = 0$ and $P'(\alpha) = 0$ but got nowhere.

OneLight
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    After substituting $\alpha$ into $\alpha^{n-1} = -p/n$ notice you can use the rule that $(a/b)^n = a^n / b^n$. Then try to cancel things out. – Bolton Bailey Oct 11 '21 at 00:33

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If you substitute $\alpha=\frac{qn}{p(n-1)}$ into the expression $\alpha^{n-1}=-\frac{p}{n}$ then you get $${\left(\frac{qn}{p(n-1)}\right)}^{n-1}=-\frac{p}{n}$$ which you can simplify to $${\left(\frac{q}{(n-1)}\right)}^{n-1}{\left(\frac{n}{p}\right)}^{n-1}=-\frac{p}{n}$$ which can get tidied up to $${\left(\frac{q}{(n-1)}\right)}^{n-1}=-{\left(\frac{p}{n}\right)}^{n}$$ $${\left(\frac{q}{(n-1)}\right)}^{n-1}+{\left(\frac{p}{n}\right)}^{n}=0$$

hopefully that helps

Nyra
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