jmc

By Vieta's formulas, $a,b,c$ are the roots of the polynomial $$x^3-3x^2+3x-3=0.$$Adding $2$ to both sides, we can factor this equation as $$(x-1{)}^3=2.$$For the real value $x=a$, we have $a-1=\sqrt[3]{2}$, so $a=\boxed{1+\sqrt[3]{2}}$.