A quadratic equation whose roots are ${\left(\frac{\gamma }{\alpha }\right)}^2$  and ${\left(\frac{\beta }{\alpha }\right)}^2$ , where $\alpha ,\beta ,\gamma$  are roots of $x^3+27=0$  is

Given that, $x^3+27=0$

$\Rightarrow x^3={\left(-3\right)}^3$

$\Rightarrow x=-3,-3\omega ,-3{\omega }^2$

$\therefore$  $\frac{\gamma }{\alpha }=\frac{-3{\omega }^2}{-3}\Rightarrow \frac{\gamma }{\alpha }={\omega }^2,$

And $\frac{\text{β}}{\alpha }=\frac{-3\omega }{-3}\Rightarrow \frac{\text{β}}{\alpha }=\omega$

$\therefore$  ${\left(\frac{\gamma }{\alpha }\right)}^2={\omega }^4=\omega$  and ${\left(\frac{\text{β}}{\alpha }\right)}^2={\omega }^2$

sum of the roots : ${\left(\frac{\gamma }{\alpha }\right)}^2$ +${\left(\frac{\beta }{\alpha }\right)}^2$  $=\omega +{\omega }^2$

product of the roots : ${\left(\frac{\gamma }{\alpha }\right)}^2$  ${\left(\frac{\beta }{\alpha }\right)}^2$ $={\omega }^3$

$\therefore$  The required equation is $x^2-\left({\left(\frac{\gamma }{\alpha }\right)}^2+{\left(\frac{\beta }{\alpha }\right)}^2\right)x+{\left(\frac{\gamma }{\alpha }\right)}^2{\left(\frac{\beta }{\alpha }\right)}^2=0$

$\Rightarrow x^2-\left(\omega +{\omega }^2\right)x+{\omega }^3=0$

$\Rightarrow x^2+x+1=0$ .