$\alpha ,\beta \text{ are the roots of the equation }{x}^2-2x+3=0\text{ . Then the equation whose roots are }$$P={\alpha }^3-3{\alpha }^2+5\alpha -2\text{ and }Q={\beta }^3-{\beta }^2+\beta +5\text{ is }$

 

$\text{ Given }\alpha ,\beta \text{ are roots of equation }{x}^2-2x+3=0$

$\Rightarrow {\alpha }^2-2\alpha +3=0 -----\left(1\right)$

$3=-{\alpha }^2+2\alpha -----\left(2\right)$

from eq1

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\alpha }^2=2\alpha -3 \\ multiply by \alpha we get {\alpha }^3=2{\alpha }^2-3\alpha \\ sub from eq 2\\ \therefore P\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\left(2{\alpha }^2-3\alpha \right)-3{\alpha }^2+5\alpha -2\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-{\alpha }^2+2\alpha -2=3-2=1 \end{array}$

$\text{ and }{\beta }^2-2\beta +3=0 ----\left(3\right)$

${\beta }^3-2{\beta }^2+3\beta =0$

${\beta }^3=2{\beta }^2-3\beta ----\left(4\right)$

$Similarly Q={\beta }^3-{\beta }^2+\beta +5$

sub from eq 4

$Q=2{\beta }^2-3\beta -{\beta }^2+\beta +5$

$={\beta }^2-2\beta +5$

=-3+5=2  from eq 3

P=1 , Q=2
So, sum of roots = 3, and product of roots = 2.

$\text{ Then required equation is }{x}^2-3x+2=0\text{ . }$