Fun With Quadratic Roots.....

By Vieta's formulas, we have: $\alpha+\beta=4$ $\alpha\beta=5$

Now, we have to express the expression $\alpha^{3}+\alpha^{4}\beta^{2}+\alpha^{2}\beta^{4}+\beta^{3}$ only with the two formulas we obtained at the beginning.

Look at the first and last terms that are $\alpha^{3}+\beta^{3}$ , they can be written as: $(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)$

Then, the terms $\alpha^{4}\beta^{2}+\alpha^{2}\beta^{4}$ can be written as: $(\alpha\beta)^{2}(\alpha^{2}+\beta^{2})$ And again, that can be written as: $(\alpha\beta)^{2}((\alpha+\beta)^{2}-2\alpha\beta)$

So, the original expression is: $(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta)+(\alpha\beta)^{2}((\alpha+\beta)^{2}-2\alpha\beta)$

Now, we replace the values of Vieta's formula: $(4)^{3}-3(5)(4)+(5)^{2}((4)^{2}-2(5))=$ $64-60+25(16-10)=$ $4+25(6)=$ $4+150=$ $\boxed{154}$