jmc

Note that $x=0$ cannot be a real root. Dividing by $x^3,$ we get $$x^3+3a{x}^2+(3a^2+3)x+a^3+6a+\frac{3a^2+3}{x}+\frac{3a}{x^2}+\frac{1}{x^3}=0.$$Let $y=x+\frac{1}{x}.$ Then $$y^2=x^2+2+\frac{1}{x^2},$$so $x^2+\frac{1}{x^2}=y^2-2,$ and $$y^3=x^3+3x+\frac{3}{x}+\frac{1}{x^3},$$so $x^3+\frac{1}{x^3}=y^3-3y.$ Thus, $$y^3-3y+3a(y^2-2)+(3a^2+3)y+a^3+6a=0.$$Simplifying, we get $$y^3+3a{y}^2+3a^{2}y+a^3=0,$$so $(y+a{)}^3=0.$ Then $y+a=0,$ so $$x+\frac{1}{x}+a=0.$$Hence, $x^2+ax+1=0.$ For the quadratic to have real roots, the discriminant must be nonnegative, so $a^2\ge 4.$ The smallest positive real number $a$ that satisfies this inequality is $a=\boxed{2}.$