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

Given, $\mathrm{\alpha }, \mathrm{\beta }$ are roots of equation

${\mathrm{x}}^2-2\mathrm{x}+3=0$

$\begin{array}{l}\Rightarrow {\mathrm{\alpha }}^2-2\mathrm{\alpha }+3=0 \left(1\right)\\ \text{ and }{\mathrm{\beta }}^2-2\mathrm{\beta }+3=0 \left(2\right) \end{array}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{\alpha }}^2=2\mathrm{\alpha }-3\text{ or }{\mathrm{\alpha }}^3=2{\mathrm{\alpha }}^2-3\mathrm{\alpha }\\ \Rightarrow \mathrm{P}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\left(2{\mathrm{\alpha }}^2-3\mathrm{\alpha }\right)-3{\mathrm{\alpha }}^2+5\mathrm{\alpha }-2\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} =-{\mathrm{\alpha }}^2+2\mathrm{\alpha }-2=3-2=1 [\mathrm{Using} (1\left)\right]\end{array}$

Similarly, we have Q = 2.
Now, sum of roots is 3 and product of roots is 2. Hence, the
required equation is ${\mathrm{x}}^2-3\mathrm{x}+2=0$