Let $\mathrm{\alpha }\ne 1$ be a real root of the equation
${\mathrm{x}}^3-{\mathrm{ax}}^2+\mathrm{ax}-1=0$ where $\mathrm{\alpha }\ne -1$ is a real number. Then a root of this equation, among the following is 

$\begin{array}{l}Letf\left(x\right)=x^3-a{x}^2+ax-1=0\\ \because \alpha isaroot\Rightarrow f\left(\alpha \right)=0\\ \Rightarrow {\alpha }^3-a{\alpha }^2+a\alpha -1=0\\ \Rightarrow {\alpha }^3(1-a\left(\frac{1}{\alpha }\right)+\frac{a}{{\alpha }^2}-\frac{1}{{\alpha }^3})=0\\ \Rightarrow 1-\frac{a}{\alpha }+\frac{a}{{\alpha }^2}-\frac{1}{{\alpha }^3}=0\\ \Rightarrow \frac{1}{{\alpha }^3}-\frac{a}{{\alpha }^2}+\frac{a}{\alpha }-1=0\\ \Rightarrow {\left(\frac{1}{\alpha }\right)}^3-a{\left(\frac{1}{\alpha }\right)}^2+a\left(\frac{1}{\alpha }\right)-1=0\\ Itisoftheformf\left(\frac{1}{\alpha }\right)=0\\ \therefore \frac{1}{\alpha }isrootoff\left(x\right)\end{array}$