If α and β are the zeros of the quadratic polynomial $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{t}}^2-4\mathrm{t}+3$, find the value of  ${\mathrm{\alpha }}^4{\mathrm{\beta }}^3+{\mathrm{\alpha }}^3{\mathrm{\beta }}^4.$

Given:  $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{t}}^2-4\mathrm{t}+3$

The roots of the equation are α and β.

$\therefore \mathrm{\alpha }+\mathrm{\beta }=4, \mathrm{\alpha \beta }=3$

Now,

${\mathrm{\alpha }}^4{\mathrm{\beta }}^3+{\mathrm{\alpha }}^3{\mathrm{\beta }}^4$

$={\mathrm{\alpha }}^3{\mathrm{\beta }}^3\left(\mathrm{\alpha }+\mathrm{\beta }\right)$

$={\left(\mathrm{\alpha \beta }\right)}^3\left(\mathrm{\alpha }+\mathrm{\beta }\right)$

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

$=27\times 4$

$=108$

Hence, ${\mathrm{\alpha }}^4{\mathrm{\beta }}^3+{\mathrm{\alpha }}^3{\mathrm{\beta }}^4=\mathbf{108}$