If the equation whose roots are the squares of the roots of the cubic $x^3-a{x}^2+bx-1=0$  is identical with the given cubic equation, then

Given equation is$x^3-a{x}^2+bx-1=0.$ If roots of the equation be $\alpha ,\beta ,\gamma ,$  then

${\alpha }^2+{\beta }^2+{\gamma }^2={\left(\alpha +\beta +\gamma \right)}^2-2\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)$

                    $=a^2-2b$

${\alpha }^2{\beta }^2+{\gamma }^2{\beta }^2+{\gamma }^2{\alpha }^2={\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)}^2-2\alpha \beta \gamma \left(\alpha +\beta +\gamma \right)$

                                 $=b^2-2a$

                    ${\alpha }^2{\beta }^2{\gamma }^2=1$

So the equation whose roots are  is given by $x^3-\left(a^2-2b\right)x^2+\left(b^2-2a\right)x-1=0$

It is identical to  $x^3-a{x}^2+bx-1=0$

$\Rightarrow a^2-2b=a$  and $b^2-2a=b$

Eliminating $b,$ we get $\frac{{\left(a^2-a\right)}^2}{4}-2a=\frac{a^2-a}{2}$

$\Rightarrow a\left\{a{\left(a-1\right)}^2-8-2\left(a-1\right)\right\}=0$

$\Rightarrow a\left(a^3-2a^2-a-6\right)=0$

$\Rightarrow a\left(a-3\right)\left(a^2+a+2\right)=0$

$\Rightarrow a=0$  or $a=3$  or $a^2+a+2=0,$

which gives $b=0$  or $b=3$  or$b^2+b+2=0.$  

So, $a=b=0$  or $a=b=3$  or$a,b$  are roots of $x^2+x+2=0.$