jmc

Since the coefficients are real, another root is $1-i\sqrt{3}.$ By Vieta's formulas, the sum of the roots is 0, so the third root is $-2.$ Hence, the cubic polynomial is $$\begin{array}{rl}(x-1-i\sqrt{3})(x-1+i\sqrt{3})(x+2)& =((x-1{)}^2-(i\sqrt{3}{)}^2)(x+2)\\ & =((x-1{)}^2+3)(x+2)\\ & =x^3+8.\end{array}$$Thus, $a=0$ and $b=8,$ so $a+b=\boxed{8}.$