jmc

Let the roots of $x^3+Ax+10$ be $p$, $q$, and $r$, and let the roots of $x^3+B{x}^2+50=0$ be $p$, $q$, and $s$. By Vieta's formulas, $$\begin{array}{rl}p+q+r& =0,\\ pqr& =-10,\\ pq+ps+qs& =0,\\ pqs& =-50.\end{array}$$From the equation $p+q+r=0,$ we conclude that $ps+qs+rs=0.$ Subtracting the equation $pq+ps+qs=0,$ we get $pq-rs=0,$ so $pq=rs.$

Then $$(pq{)}^3=(pq)(pq)(rs)=(pqr)(pqs)=(-10)(-50)=500.$$Therefore, $pq=\sqrt[3]{500}=5\sqrt[3]{4}$. The final answer is $5+3+4=\boxed{12}.$