If $\mathrm{\omega }$ is an imaginary cube root of unity then the equation whose roots are $2\mathrm{\omega }+3{\mathrm{\omega }}^2\text{ and }2{\mathrm{\omega }}^2+3\mathrm{\omega }$ is

Given that $2\mathrm{\omega }+3{\mathrm{\omega }}^2\text{ and }2{\mathrm{\omega }}^2+3\mathrm{\omega }$ are the roots of the equation

here $\omega$ is complex cube root of unity, so that $1+\omega +{\omega }^2=0 and {\omega }^3=1$

The equation whose roots are $2\mathrm{\omega }+3{\mathrm{\omega }}^2\text{ and }2{\mathrm{\omega }}^2+3\mathrm{\omega }$ is     

                $\begin{array}{rcl}{x}^2-x\left(2\mathrm{\omega }+3{\mathrm{\omega }}^2+2{\mathrm{\omega }}^2+3\mathrm{\omega }\right)+\left(2\mathrm{\omega }+3{\mathrm{\omega }}^2\right)·\left(2{\mathrm{\omega }}^2+3\mathrm{\omega }\right)& =& 0\\ x^2-x\left(5\right)\left(\omega +{\omega }^2\right)+4+6{\omega }^2+6\omega +9& =& 0\\ x^2-5x\left(-1\right)+13+6\left(-1\right)& =& 0\\ x^2+5x+7& =& 0\end{array}$

Therefore, the equation whose roots are $2\mathrm{\omega }+3{\mathrm{\omega }}^2\text{ and }2{\mathrm{\omega }}^2+3\mathrm{\omega }$ is $x^2+5x+7=0$