If $\mathrm{\alpha },\mathrm{\beta }$ are two real numbers satisfying ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=5\text{ and }3\left({\mathrm{\alpha }}^5+{\mathrm{\beta }}^5\right)=11\left({\mathrm{\alpha }}^3+{\mathrm{\beta }}^3\right)$ then $\mathrm{\alpha \beta }$ is 

$$\begin{gathered} {\alpha }^2+{\beta }^2=5\text{ and }3\left({\alpha }^5+{\beta }^5\right)=11\left({\alpha }^3+{\beta }^3\right) \\ 3\left({\alpha }^5+{\beta }^5\right)=11\left({\alpha }^3+{\beta }^3\right) \\ 3(\alpha +\beta )({\alpha }^4-{\alpha }^3\beta +{\alpha }^2{\beta }^2-\alpha {\beta }^3+{\beta }^4)=11(\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2) \\ 3\left[{({\alpha }^2+{\beta }^2)}^2-{\alpha }^2{\beta }^2-\alpha \beta ({\alpha }^2+{\beta }^2)\right]=11\left({\alpha }^2+{\beta }^2-\alpha \beta \right) \\ 3(25-{\alpha }^2{\beta }^2-5\alpha \beta )=11(5-\alpha \beta ) \\ 75-3{\left(\alpha \beta \right)}^2-15\alpha \beta =55-11\alpha \beta \\ 3{\left(\alpha \beta \right)}^2+15\alpha \beta -20=0 \\ (\alpha \beta -2)(3\alpha \beta +10)=0 \\ \alpha \beta =2 \\ Multiplying both \\ 11\left({\alpha }^3+{\beta }^3\right)({\alpha }^2+{\beta }^2)=15\left({\alpha }^5+{\beta }^5\right) \\ 11({\alpha }^5+{\beta }^5+{\alpha }^2{\beta }^2(\alpha +\beta \left)\right)=15\left({\alpha }^5+{\beta }^5\right) \end{gathered}$$