jmc

by Vieta's formulas, $a+b=4$ and $ab=5.$ Then $$\begin{array}{rl}{a}^3+b^3& =(a+b)(a^2-ab+b^2)\\ & =(a+b)(a^2+2ab+b^2-3ab)\\ & =(a+b)((a+b{)}^2-3ab)\\ & =4\cdot (4^2-3\cdot 5)\\ & =4,\end{array}$$and $$\begin{array}{rl}{a}^4{b}^2+a^2{b}^4& =a^2{b}^2(a^2+b^2)\\ & =(ab{)}^2((a+b{)}^2-2ab)\\ & =5^2(4^2-2\cdot 5)\\ & =150,\end{array}$$so $a^3+a^4{b}^2+a^2{b}^4+b^3=\boxed{154}.$