let $z_1,z_2,z_3,..,z_n\in\mathbb{R}^{+}$ be the root of $ax^n-ax^{n-1}+a_2x^{n-2}+\ldots+a_{n-2}x^2-n^2bx+b=0$ there for $\sum_{k=1}^{n} z_k=-\frac{-a}{a}=1~~~(*),~~~~$ $~~~~\prod_{k=1}^n z_k=(-1)^n\frac{b}{a}~~~(**)$ and $\prod_{k=1}^n z_k(\sum_{k=1}^{n} \frac{1}{z_k})=(-1)^{n}\frac{n^2b}{a}$ thus $\sum_{k=1}^{n} \frac{1}{z_k}=n^2~~~(***)~~~;~from~(**)$ from $AM\geq HM: ~\frac{1}{n} \sum_{k=1}^{n} z_k \geq \frac{n}{ \sum_{k=1}^{n} \frac{1}{z_k}}$ which the equality case holds at $z_1=z_2=z_3=...=z_n$ from $(*),~(***)$ we'll get that $\frac{1}{n} \sum_{k=1}^{n} z_k=\frac{1}{n}=\frac{n}{ \sum_{k=1}^{n} \frac{1}{z_k}}$ $\therefore$ we can conclude that $z_1=z_2=z_3=...=z_n$ as desired $\square$