jmc

Expanding, we get $$(a+b{)}^2+(b+c{)}^2+(c+a{)}^2=2(a^2+b^2+c^2)+2(ab+bc+ca).$$To compute this expression, note that $$(a+b+c{)}^2=(a^2+b^2+c^2)+2(ab+bc+ca).$$Then we can write the given expression in terms of $a+b+c$ and $ab+bc+ca$: $$\begin{array}{rl}2(a^2+b^2+c^2)+2(ab+bc+ca)& =[2(a^2+b^2+c^2)+4(ab+bc+ca)]-2(ab+bc+ca)\\ & =2(a+b+c{)}^2-2(ab+bc+ca).\end{array}$$By Vieta's formulas, $a+b+c=20$ and $ab+bc+ca=18$, so the answer is $2\cdot 20^2-2\cdot 18=\boxed{764}.$