Description for Correct answer:
Let \( \Large a, a^{n} \) be the roots of equation \( \Large ax^{2}+bx+c=0 \)

then \( \Large a+a^{n}=\frac{-b}{a} and\ a.a^{n}=\frac{c}{a} => a^{n+1}=\frac{c}{a} \)

Eliminating \( \Large \alpha \), we get

\( \Large \left(\frac{c}{a}\right)^{\frac{1}{n+1}}+ \left(\frac{c}{a}\right)^{\frac{n}{n+1}} =-\frac{b}{a} \)

=> \( \Large a.a^{-\frac{1}{n+1}\frac{1}{c^{n+1}}}+a.a^{-\frac{n}{n+1}\frac{n}{n+1}}=-b \)

=> \( \Large \left(a^{n}c\right)^{\frac{1}{n+1}}+ \left(ac^{n}\right)^{\frac{1}{n+1}}=-b \)