If the cubic equation ${\mathrm{x}}^3-{\mathrm{ax}}^2+\mathrm{ax}-1=0$ is identical with the cubic equation whose roots are the squares of the roots of the given cubic equation, then the non zero real value of a is

$\begin{array}{l}Let\alpha ,\beta ,\gamma arerootsof{x}^3-a{x}^2+ax-1=0--\left(1\right)\\ \therefore \alpha +\beta +\gamma =a,\alpha \beta +\beta \gamma +\gamma \alpha =a,\alpha \beta \gamma =1\\ TheEquationhaving{\alpha }^2,{\beta }^2,{\gamma }^{2}arerootsis\\ x^3-({\alpha }^2+{\beta }^2+{\gamma }^2)x^2+({\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2)x-{\alpha }^2{\beta }^2{\gamma }^2=0--\left(2\right)\\ \left(1\right)and\left(2\right)areidentical\\ \therefore \frac{-({\alpha }^2+{\beta }^2+{\gamma }^2)}{-a}=\frac{{\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2}{a}=\frac{-{\alpha }^2{\beta }^2{\gamma }^2}{-1}\\ \Rightarrow \frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{a}=\frac{{\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2}{a}=1[\because \alpha \beta \gamma =1]\\ \therefore {\alpha }^2+{\beta }^2+{\gamma }^2=a\\ \Rightarrow (\alpha +\beta +\gamma )-2(\alpha \beta +\beta \gamma +\gamma \alpha )=a\\ \Rightarrow a^2-2a=a\\ \Rightarrow a^2-3a=0\\ \Rightarrow a(a-3)=0\\ \Rightarrow a=3\end{array}$