If one root of ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ is equal to ${\mathrm{n}}^{\mathrm{th}}$ power of the other, then

${\left({\mathrm{ac}}^{\mathrm{n}}\right)}^{1/\left(\mathrm{n}+1\right)}+{\left({\mathrm{a}}^{\mathrm{n}}\mathrm{c}\right)}^{1/\left(\mathrm{n}+1\right)}=$

$\begin{array}{l}Let\alpha ,{\alpha }^{n}arerootsofa{x}^2+bx+c=0\\ Then\alpha .{\alpha }^n=\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\Rightarrow {\alpha }^{n+1}=\raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\\ and\alpha +{\alpha }^n=\frac{-b}{a}\\ \Rightarrow {\left(\frac{c}{a}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.+1}+{\left(\frac{c}{a}\right)}^{\frac{n}{n+1}}=\frac{-b}{a}\\ \Rightarrow a{\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}+a{\left(\frac{c}{a}\right)}^{\frac{n}{n+1}}=-b\\ \Rightarrow {\left(a^{n}c\right)}^{\frac{1}{n+1}}+{\left(a{c}^n\right)}^{\frac{1}{n+1}}=-b\end{array}$